An original problem · started in Calculus 2 and Linear Algebra
I came up with this problem as a student, while I was taking Calculus 2, where geometric series live, and a computational linear algebra course, where rotation matrices live. The idea was to blend the two: what happens to a geometric series if every term is an arrow, and every arrow is turned a little from the one before it?
Start with a series everybody knows:
$$\sum_{n=1}^{\infty} \left(\tfrac{1}{2}\right)^{n-1} = 1 + \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \cdots = 2.$$Usually you think of it as a sequence of partial sums of real numbers. Instead, think of it as a sequence of partial sums of vectors in $\mathbb{R}^2$: an arrow of length $1$, then an arrow of length $\frac{1}{2}$ placed at its tip, then one of length $\frac{1}{4}$, and so on. Lay them all along the positive $x$-axis and the tip of the chain closes in on the point $(2, 0)$.
Choosing the positive $x$-axis as the "first" direction is arbitrary, but it stays the standard for the rest of this page. To write the arrows down, we need the matrix that turns a vector.
For an angle $\theta$, the rotation matrix
$$R(\theta) = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$$turns every vector in $\mathbb{R}^2$ counterclockwise by $\theta$ radians about the origin.
With a bit of linear algebra, the series of arrows along the $x$-axis can be written as
$$\sum_{n=1}^{\infty} \left(\tfrac{1}{2}\right)^{n-1} R(0)^{n-1} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \end{bmatrix}.$$Here $R(0)$ is the identity matrix, so the rotation looks completely superfluous. Its job becomes clear in a moment. First, one fact about powers of rotations.
For every angle $\theta$ and every integer $k$,
$$R(\theta)^k = R(k\theta) = \begin{bmatrix} \cos(k\theta) & -\sin(k\theta) \\ \sin(k\theta) & \cos(k\theta) \end{bmatrix}.$$Multiplying out and using the angle-addition formulas gives $R(\alpha) R(\beta) = R(\alpha + \beta)$. With $R(0) = I$, induction gives $R(\theta)^k = R(k\theta)$ for $k \geq 0$. Since $R(\theta) R(-\theta) = R(0) = I$, the inverse of $R(\theta)$ is $R(-\theta)$, so for $k < 0$ we get $R(\theta)^k = R(-\theta)^{-k} = R(k\theta)$ as well. $\blacksquare$
Turning $k$ times by $\theta$ is the same as turning once by $k\theta$. That's what makes the next example work.
Find the sum of the alternating series $1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \frac{1}{32} + \cdots$, first as real numbers and then as vectors in $\mathbb{R}^2$.
As real numbers, split it into its positive and negative terms:
$$\left(1 + \tfrac{1}{4} + \tfrac{1}{16} + \cdots\right) - \left(\tfrac{1}{2} + \tfrac{1}{8} + \tfrac{1}{32} + \cdots\right) = \frac{1}{1 - \frac{1}{4}} - \frac{\frac{1}{2}}{1 - \frac{1}{4}} = \frac{4}{3} - \frac{2}{3} = \frac{2}{3}.$$As vectors, a minus sign is a half turn: each new arrow points the opposite way from the one before it, which is a rotation by $\pi$ radians. By the identity,
$$\sum_{n=1}^{\infty} \left(\tfrac{1}{2}\right)^{n-1} R(\pi)^{n-1} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \sum_{n=1}^{\infty} \left(\tfrac{1}{2}\right)^{n-1} \begin{bmatrix} \cos((n-1)\pi) \\ \sin((n-1)\pi) \end{bmatrix} = \sum_{n=1}^{\infty} \begin{bmatrix} \left(-\frac{1}{2}\right)^{n-1} \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} \\ 0 \end{bmatrix}.$$Same answer, of course: the chain zigzags back and forth along the $x$-axis and settles at $\left(\frac{2}{3}, 0\right)$. The real-number way was easier here. The vector way is the one that keeps working when the turn isn't a half turn.
Here are both ways at once.
We went through this calculus gymnastics to consider the following: the same geometric series, but each successive arrow turned counterclockwise by $\frac{\pi}{4}$ radians. The chain no longer stays on a line. It spirals inward, turning an eighth of a circle at every step. There are certainly many different ways to find where it ends up, and you're encouraged to try one on your own before reading on. (There's an exercise for it below.)
Every example so far has the same shape: a sequence of lengths, and a fixed turn between consecutive arrows. That deserves a name.
For a sequence $(a_n)_{n \geq 1}$ and an angle $\theta$, the sequence $a_n$ bent by $\theta$ is the series
$$\sum_{n=1}^{\infty} a_n\, R(\theta)^{n-1} \begin{bmatrix} 1 \\ 0 \end{bmatrix}.$$Here's what the bend does, one arrow at a time: each new arrow is the last one, turned by $\theta$ and scaled to the next term. For $\left(\frac{1}{2}\right)^{n-1}$ bent by $\frac{\pi}{4}$:
Again, the starting vector $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ is an arbitrary choice. Bending by $\theta = 0$ gives back the ordinary series laid along the $x$-axis, and bending by $\theta = \pi$ gives the alternating series. Try other angles and ratios below.
When $a_n$ is geometric and the turn goes evenly into a full circle, the infinite bent series collapses to a finite sum.
Let $0 < a < b$, let $K \geq 2$ be an integer, and let $\theta = \frac{2\pi}{K}$, so that $K\theta = 2\pi$. Then the sequence $\left(\frac{a}{b}\right)^{n-1}$ bent by $\theta$ converges, and
$$\sum_{n=1}^{\infty} \left(\tfrac{a}{b}\right)^{n-1} R(\theta)^{n-1} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \frac{\displaystyle\sum_{n=1}^{K} \left(\tfrac{a}{b}\right)^{n-1} R(\theta)^{n-1} \begin{bmatrix} 1 \\ 0 \end{bmatrix}}{1 - \left(\frac{a}{b}\right)^{K}}.$$Write $r = \frac{a}{b}$, $v = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$, and $A_m = R(\theta)^m$. By the identity, the $n$-th term is $r^{n-1} \begin{bmatrix} \cos((n-1)\theta) \\ \sin((n-1)\theta) \end{bmatrix}$. Each entry is at most $r^{n-1}$ in absolute value, and $\sum r^{n-1}$ converges because $0 < r < 1$, so both coordinate series converge absolutely and the limit
$$A = \sum_{n=1}^{\infty} r^{n-1} A_{n-1} v$$exists. Since $K\theta = 2\pi$, we have $A_K = R(2\pi) = I = A_0$, and so $A_{m+K} = A_m$ for every $m \geq 0$: after $K$ turns, the arrows point the same way again. Split off the first $K$ terms and reindex the rest with $n = m + K$:
$$A = \sum_{n=1}^{K} r^{n-1} A_{n-1} v + \sum_{m=1}^{\infty} r^{m-1+K} A_{m-1+K} v = \sum_{n=1}^{K} r^{n-1} A_{n-1} v + r^K \sum_{m=1}^{\infty} r^{m-1} A_{m-1} v = \sum_{n=1}^{K} r^{n-1} A_{n-1} v + r^K A.$$Solving for $A$ gives $A = \dfrac{\sum_{n=1}^{K} r^{n-1} A_{n-1} v}{1 - r^K}$, since $r^K \neq 1$. $\blacksquare$
The idea of the proof: one full turn of $K$ arrows brings the direction back to where it started, so the rest of the chain is a copy of the whole chain, shrunk by $\left(\frac{a}{b}\right)^K$. It's the same trick that sums an ordinary geometric series, done one full turn at a time.
Here's the proof as a picture, with $\frac{a}{b} = \frac{3}{4}$ and $\theta = \frac{\pi}{2}$, so $K = 4$.
Now the spiral from earlier is a one-liner.
Find the sum of $\left(\frac{1}{2}\right)^{n-1}$ bent by $\theta = \frac{\pi}{4}$.
Here $\frac{a}{b} = \frac{1}{2}$ and $K = 8$. One full turn is the first eight arrows, the $n$-th one being $\left(\frac{1}{2}\right)^{n-1} \begin{bmatrix} \cos((n-1)\frac{\pi}{4}) \\ \sin((n-1)\frac{\pi}{4}) \end{bmatrix}$. The arrows at angles $0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$ give whole numbers, and the diagonal ones each carry a $\frac{\sqrt{2}}{2}$. Adding them up:
$$\sum_{n=1}^{8} \left(\tfrac{1}{2}\right)^{n-1} R\!\left(\tfrac{\pi}{4}\right)^{n-1} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 - \frac{1}{16} + \sqrt{2}\left(\frac{1}{4} - \frac{1}{16} - \frac{1}{64} + \frac{1}{256}\right) \\[4pt] \frac{1}{4} - \frac{1}{64} + \sqrt{2}\left(\frac{1}{4} + \frac{1}{16} - \frac{1}{64} - \frac{1}{256}\right) \end{bmatrix} = \begin{bmatrix} \frac{15}{16} + \frac{45}{256}\sqrt{2} \\[4pt] \frac{15}{64} + \frac{75}{256}\sqrt{2} \end{bmatrix}.$$Dividing by $1 - \left(\frac{1}{2}\right)^8 = \frac{255}{256}$ gives
$$\sum_{n=1}^{\infty} \left(\tfrac{1}{2}\right)^{n-1} R\!\left(\tfrac{\pi}{4}\right)^{n-1} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{16}{17} + \frac{3}{17}\sqrt{2} \\[4pt] \frac{4}{17} + \frac{5}{17}\sqrt{2} \end{bmatrix} \approx \begin{bmatrix} 1.191 \\ 0.651 \end{bmatrix}.$$Check it in the widget: set $\theta = \frac{\pi}{4}$ and $r = 0.5$, and the ring sits at about $(1.191, 0.651)$.
So far the angle has been fixed and the sum was the prize. Now flip it around: let the angle vary. Plug in an angle, and out comes a point of the plane.
For the sequence $\left(\frac{1}{2}\right)^{n-1}$, the Bender is the function $B \colon \mathbb{R} \to \mathbb{R}^2$ given by
$$B(\theta) = \sum_{n=1}^{\infty} \left(\tfrac{1}{2}\right)^{n-1} R(\theta)^{n-1} \begin{bmatrix} 1 \\ 0 \end{bmatrix}.$$It's defined for every $\theta$, not just the ones that go evenly into a full turn: each entry of the $n$-th term is at most $\left(\frac{1}{2}\right)^{n-1}$ in absolute value, so the same comparison as in the theorem's proof shows the series converges. Since $R(\theta + 2\pi) = R(\theta)$, the Bender only cares about $\theta$ up to full turns.
My question as an undergrad was: what's the image of this function? Every angle gives a point. What shape do all those points make? We already know three of them: $B(0) = (2, 0)$, $B(\pi) = \left(\frac{2}{3}, 0\right)$, and $B\left(\frac{\pi}{4}\right) = \left(\frac{16}{17} + \frac{3}{17}\sqrt{2},\ \frac{4}{17} + \frac{5}{17}\sqrt{2}\right)$. It took me well over two years of on-and-off thinking to pin down the answer.
The key is the same trick as the theorem, with $K = 1$: peel off just the first arrow. Everything after it is the whole chain again, turned by $\theta$ and shrunk by half.
The picture says $B(\theta) = \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \frac{1}{2} R(\theta)\, B(\theta)$, and since turning a vector never changes its length, the distance from $B(\theta)$ to $(1, 0)$ is always exactly half its distance to the origin. Now let $\theta$ run all the way around.
The image of the Bender is the circle with center $\left(\frac{4}{3}, 0\right)$ and radius $\frac{2}{3}$. As $\theta$ runs over $(-\pi, \pi]$, the point $B(\theta)$ goes around this circle exactly once.
Write $r = \frac{1}{2}$ and $v = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$, and fix $\theta$.
Step 1: peel off the first arrow. Splitting off the $n = 1$ term and reindexing with $n = m + 1$,
$$B(\theta) = v + \sum_{m=1}^{\infty} r^{m} R(\theta)^{m} v = v + r R(\theta) \sum_{m=1}^{\infty} r^{m-1} R(\theta)^{m-1} v = v + r R(\theta)\, B(\theta).$$Pulling $r R(\theta)$ out of the series is allowed because it's linear: it comes out of every partial sum, and so out of the limit.
Step 2: rotations keep lengths. Rearranging, $B(\theta) - v = r R(\theta) B(\theta)$. Since $\lVert R(\theta) w \rVert = \lVert w \rVert$ for every vector $w$,
$$\lVert B(\theta) - v \rVert = r\, \lVert B(\theta) \rVert.$$Step 3: that's a circle. Write $B(\theta) = (x, y)$ and square both sides of Step 2:
$$(x - 1)^2 + y^2 = r^2 (x^2 + y^2) \iff (1 - r^2)(x^2 + y^2) - 2x + 1 = 0 \iff \left(x - \frac{1}{1 - r^2}\right)^2 + y^2 = \left(\frac{r}{1 - r^2}\right)^2.$$With $r = \frac{1}{2}$, that's the circle with center $\left(\frac{4}{3}, 0\right)$ and radius $\frac{2}{3}$. So every $B(\theta)$ lies on it.
Step 4: every point of the circle is hit, exactly once. Take any point $P$ on the circle. The origin is not on the circle, so $P \neq 0$, and Step 3 run backward gives $\lVert P - v \rVert = r \lVert P \rVert$. So $P - v$ and $rP$ are nonzero vectors of the same length, and exactly one $\theta \in (-\pi, \pi]$ turns one onto the other: $r R(\theta) P = P - v$, that is, $P = v + r R(\theta) P$. Now $B(\theta)$ solves the same equation $X = v + r R(\theta) X$, and that equation has only one solution, because the matrix $I - r R(\theta)$ is invertible:
$$\det\big(I - r R(\theta)\big) = (1 - r\cos\theta)^2 + r^2 \sin^2\theta = (1 - r)^2 + 2r(1 - \cos\theta) > 0.$$So $P = B(\theta)$. Every point of the circle comes from some $\theta$, and from only one $\theta$ in $(-\pi, \pi]$, since that $\theta$ was the only one that worked. $\blacksquare$
A quick check: $B(0) = (2, 0)$ and $B(\pi) = \left(\frac{2}{3}, 0\right)$ are the two ends of a horizontal diameter, and their midpoint is $\frac{4}{3}$. The $\frac{\pi}{4}$ sum lands on the circle too.
Notice that the proof never used $r = \frac{1}{2}$ until the very last line. For any ratio $0 < r < 1$, the image of $r^{n-1}$ bent by $\theta$ is the circle with center $\frac{1}{1 - r^2}$ on the $x$-axis and radius $\frac{r}{1 - r^2}$, running from $\frac{1}{1 + r}$ (at $\theta = \pi$) to $\frac{1}{1 - r}$ (at $\theta = 0$). The Bend a Series widget above lets you test it: fix $r$, slide $\theta$, and watch the ring stay on one circle.
Every geometric sequence that goes to zero gives a circle; only the center and radius change. So here's the fateful question: what about other sequences that go to zero? Are there images besides circles?
For any sequence $(a_n)$, write $B_a(\theta) = \sum_{n=1}^{\infty} a_n R(\theta)^{n-1} \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ for its Bender. Before hunting for shapes, it helps to know when the hunt is safe. If the lengths of the arrows add up to something finite, everything behaves.
Let $(a_n)$ be a sequence of real numbers with $\sum_{n=1}^{\infty} |a_n|$ convergent. Then $B_a(\theta)$ converges for every $\theta$, the function $B_a$ is continuous, $B_a(\theta + 2\pi) = B_a(\theta)$, and $B_a(-\theta)$ is the mirror image of $B_a(\theta)$ across the $x$-axis. So the image of $B_a$ is a closed curve, symmetric across the $x$-axis.
By the identity, the $n$-th term is $a_n \begin{bmatrix} \cos((n-1)\theta) \\ \sin((n-1)\theta) \end{bmatrix}$, and each entry is at most $|a_n|$ in absolute value. Since $\sum |a_n|$ converges, both coordinate series converge absolutely, for every $\theta$.
For continuity: after the first $N$ terms, each coordinate of what's left is at most $\sum_{n > N} |a_n|$ in absolute value, for every $\theta$ at once. That bound goes to $0$ as $N \to \infty$, so the partial sums, which are continuous functions of $\theta$, converge to $B_a$ uniformly (the Weierstrass M-test). A uniform limit of continuous functions is continuous.
Periodicity comes from $R(\theta + 2\pi) = R(\theta)$. For the mirror image, $\cos(-(n-1)\theta) = \cos((n-1)\theta)$ and $\sin(-(n-1)\theta) = -\sin((n-1)\theta)$ term by term, so $B_a(-\theta)$ has the same $x$-coordinate as $B_a(\theta)$ and the opposite $y$-coordinate. $\blacksquare$
The symmetry gives a quick way to prove an image is not a circle, instead of just eyeballing it.
Let $(a_n)$ be as in the lemma, with $B_a(0) \neq B_a(\pi)$. If the image of $B_a$ is a circle, then it is the circle whose diameter is the segment from $B_a(\pi)$ to $B_a(0)$.
$B_a(0) = \left(\sum a_n, 0\right)$ and $B_a(\pi) = \left(\sum (-1)^{n-1} a_n, 0\right)$ both lie on the $x$-axis. Suppose the image is a circle. By the lemma, reflecting across the $x$-axis maps the image onto itself, so it maps the circle onto itself, and so it fixes the circle's center: the center is on the $x$-axis. A circle centered on the $x$-axis meets the axis in exactly two points, the ends of a horizontal diameter. $B_a(0)$ and $B_a(\pi)$ are two different points of the circle on the axis, so they are those two ends. $\blacksquare$
So to rule out a circle, find one point of the image that is not on the circle through $B_a(\pi)$ and $B_a(0)$. Finite sequences count here: a sequence that's eventually zero certainly goes to zero.
Show that the image of $(1, 1, 1, 0, 0, \dots)$ bent by $\theta$ is not a circle.
Only three arrows, so $B(\theta) = \begin{bmatrix} 1 + \cos\theta + \cos 2\theta \\ \sin\theta + \sin 2\theta \end{bmatrix}$. Then $B(0) = (3, 0)$ and $B(\pi) = (1, 0)$, so by the circle test the only possible circle has center $(2, 0)$ and radius $1$. But $B\left(\frac{\pi}{2}\right) = (1 + 0 - 1,\ 1 + 0) = (0, 1)$, which is $\sqrt{5}$ from $(2, 0)$, not $1$. Not a circle.
That one is finite. Here's an infinite sequence that goes to zero, and the check needs two famous sums.
Show that the image of $\frac{1}{n^2}$ bent by $\theta$ is not a circle.
Since $\sum \frac{1}{n^2}$ converges, the lemma applies. At $\theta = 0$ every arrow points right, so $B(0) = \left(\sum \frac{1}{n^2}, 0\right) = \left(\frac{\pi^2}{6}, 0\right)$, the Basel sum. At $\theta = \pi$ the arrows alternate, and $B(\pi) = \left(\sum \frac{(-1)^{n-1}}{n^2}, 0\right) = \left(\frac{\pi^2}{12}, 0\right)$: subtracting twice the even terms, $\frac{2}{4} \cdot \frac{\pi^2}{6}$, from the Basel sum. So the only possible circle has center $\frac{\pi^2}{8} \approx 1.2337$ and radius $\frac{\pi^2}{24} \approx 0.4112$.
Now try $\theta = \frac{\pi}{2}$. The arrows point right, up, left, down, right, ..., so
$$B\left(\tfrac{\pi}{2}\right) = \left(1 - \tfrac{1}{3^2} + \tfrac{1}{5^2} - \cdots,\ \ \tfrac{1}{2^2} - \tfrac{1}{4^2} + \tfrac{1}{6^2} - \cdots\right).$$The second coordinate is $\frac{1}{4}\left(1 - \frac{1}{2^2} + \frac{1}{3^2} - \cdots\right) = \frac{1}{4} \cdot \frac{\pi^2}{12} = \frac{\pi^2}{48} \approx 0.2056$. The first is Catalan's constant $G$; it has no famous closed form, but by the alternating series estimate, its first $16$ terms pin it down to within $0.001$: $G \approx 0.9160$. The distance from $B\left(\frac{\pi}{2}\right)$ to the center is about $\sqrt{(0.9160 - 1.2337)^2 + 0.2056^2} \approx 0.378$, nowhere near $0.4112$. Not a circle: the point sits inside the test circle.
Your turn. Pick a sequence, or build your own first six terms, and watch its image draw itself. The dashed green circle is the circle test; the caption says whether the image passes.
The Bender is linear in the sequence: bending $(1 - s)\,a + s\,b$ gives $(1 - s) B_a(\theta) + s B_b(\theta)$. So you can slide continuously from one shape to the next.
So yes: there are plenty of images besides circles. Every sequence in this section had $\sum |a_n| < \infty$, so the lemma kept everything tame. The first sequence anyone reaches for that goes to zero without that, the harmonic sequence $\frac{1}{n}$, is where things get wild.
When I asked about other sequences, I went straight to the harmonic one. It turned out to have a lot of structure, and some awesome special values.
The harmonic Bender is the harmonic sequence $\frac{1}{n}$ bent by $\theta$:
$$B_h(\theta) = \sum_{n=1}^{\infty} \frac{1}{n}\, R(\theta)^{n-1} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \sum_{n=1}^{\infty} \frac{1}{n} \begin{bmatrix} \cos((n-1)\theta) \\ \sin((n-1)\theta) \end{bmatrix},$$for every $\theta$ at which the series converges.
The lemma from the last section doesn't apply: the arrows have lengths $1, \frac{1}{2}, \frac{1}{3}, \dots$, and those add up to infinity. So convergence is no longer free, and it really does fail somewhere. At $\theta = 0$ every arrow points right, the partial sums are $(H_N, 0)$ with $H_N = 1 + \frac{1}{2} + \cdots + \frac{1}{N}$, and the harmonic series diverges. The chain marches off to the right without bound, toward "$(\infty, 0)$." The same happens at every multiple of $2\pi$.
Everywhere else, though, the turning saves it. The arrows keep changing direction, so they keep partly canceling, and the partial sums of the directions stay bounded.
If $\theta$ is not a multiple of $2\pi$, then for every $m \geq 0$,
$$\left| \sum_{k=0}^{m} \cos(k\theta) \right| \leq \frac{1}{|\sin(\theta/2)|} \qquad \text{and} \qquad \left| \sum_{k=0}^{m} \sin(k\theta) \right| \leq \frac{1}{|\sin(\theta/2)|}.$$The product-to-sum formulas give $2\sin\frac{\theta}{2}\cos(k\theta) = \sin\left(\left(k + \tfrac{1}{2}\right)\theta\right) - \sin\left(\left(k - \tfrac{1}{2}\right)\theta\right)$, so summing from $k = 0$ to $m$ telescopes:
$$2\sin\tfrac{\theta}{2} \sum_{k=0}^{m} \cos(k\theta) = \sin\left(\left(m + \tfrac{1}{2}\right)\theta\right) + \sin\tfrac{\theta}{2}.$$The right side is at most $2$ in absolute value, and $\sin\frac{\theta}{2} \neq 0$, so dividing gives the first bound. The second is the same, starting from $2\sin\frac{\theta}{2}\sin(k\theta) = \cos\left(\left(k - \tfrac{1}{2}\right)\theta\right) - \cos\left(\left(k + \tfrac{1}{2}\right)\theta\right)$. $\blacksquare$
Bounded partial sums of the directions, times lengths that shrink to zero: that's exactly what the Dirichlet test needs.
The harmonic Bender $B_h(\theta)$ converges for every $\theta$ that is not a multiple of $2\pi$, and diverges at every multiple of $2\pi$. It never converges absolutely.
Divergence at multiples of $2\pi$ is the harmonic series, as above. And the arrows have lengths $\frac{1}{n}$, whose sum diverges, so the series is never absolutely convergent.
Now let $\theta$ not be a multiple of $2\pi$, and look at the $x$-coordinate $\sum_{k \geq 0} \frac{\cos(k\theta)}{k+1}$ (the $y$-coordinate is the same with sines). Write $C_m = \sum_{k=0}^{m} \cos(k\theta)$, so $|C_m| \leq K = \frac{1}{|\sin(\theta/2)|}$ by the lemma. Summing by parts, for $M < P$,
$$\sum_{k=M+1}^{P} \frac{\cos(k\theta)}{k+1} = \frac{C_P}{P+1} - \frac{C_M}{M+2} + \sum_{k=M+1}^{P-1} \left(\frac{1}{k+1} - \frac{1}{k+2}\right) C_k.$$Every coefficient $\frac{1}{k+1} - \frac{1}{k+2}$ is positive and they telescope, so the absolute value is at most $K\left(\frac{1}{P+1} + \frac{1}{M+2} + \frac{1}{M+2} - \frac{1}{P+1}\right) = \frac{2K}{M+2}$. That goes to $0$ as $M \to \infty$, no matter how large $P$ is, so the partial sums form a Cauchy sequence, and the series converges. $\blacksquare$
The proof gives more than convergence: after the first $N$ arrows, each coordinate is within $\frac{2}{(N+1)|\sin(\theta/2)|}$ of its limit. The closer $\theta$ is to $0$, the slower the chain settles down.
Since $\theta = 0$ is off limits, what I actually did was plot the image of the finite sums, the first $N$ arrows bent by every angle, and then keep pushing $N$ up.
The point at $\theta = 0$ runs away (logarithmically slowly), dragging a long thin tongue of the curve out to the right, while every other angle settles toward its limit. Try it yourself.
Now the special values: the angles where the limit can be pinned down exactly.
Find $B_h(\pi)$ and $B_h\left(\frac{\pi}{2}\right)$.
At $\theta = \pi$, the arrows alternate right and left, every sine is $0$, and
$$B_h(\pi) = \begin{bmatrix} 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots \\ 0 \end{bmatrix} = \begin{bmatrix} \ln 2 \\ 0 \end{bmatrix},$$the alternating harmonic series (the Maclaurin series of $\ln(1 + x)$ at $x = 1$).
At $\theta = \frac{\pi}{2}$, the arrows turn a quarter turn each time: right $1$, up $\frac{1}{2}$, left $\frac{1}{3}$, down $\frac{1}{4}$, right $\frac{1}{5}$, ... The horizontal arrows are the odd-numbered ones and only move $x$; the vertical arrows are the even-numbered ones and only move $y$. So
$$x = 1 + 0 - \tfrac{1}{3} + 0 + \tfrac{1}{5} + 0 - \cdots = 1 - \tfrac{1}{3} + \tfrac{1}{5} - \tfrac{1}{7} + \cdots = \frac{\pi}{4},$$Leibniz's series (the Maclaurin series of $\arctan x$ at $x = 1$), and
$$y = 0 + \tfrac{1}{2} + 0 - \tfrac{1}{4} + 0 + \tfrac{1}{6} - \cdots = \tfrac{1}{2}\left(1 - \tfrac{1}{2} + \tfrac{1}{3} - \cdots\right) = \tfrac{1}{2}\ln 2 = \ln\sqrt{2}.$$So $B_h\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{4},\ \ln\sqrt{2}\right)$. By the mirror symmetry, $B_h\left(-\frac{\pi}{2}\right) = \left(\frac{\pi}{4},\ -\ln\sqrt{2}\right)$.
Here's the quarter-turn chain feeding both famous series at once.
Dropping the zeros at $\theta = \frac{\pi}{2}$ is safe: every partial sum of $1 + 0 - \frac{1}{3} + 0 + \cdots$ equals a partial sum of $1 - \frac{1}{3} + \cdots$ (it just repeats each one twice), so the two series have the same limit. What you may not do with a series that isn't absolutely convergent is reorder its terms. Riemann's rearrangement theorem says reordering $1 - \frac{1}{2} + \frac{1}{3} - \cdots$ can make it add up to any real number you like. Which raises a question about the Bender: if you lay the same arrows down in a different order, where can the chain end up? (It's one of the open questions at the end.)
After working out $\theta = \pi$, $\frac{\pi}{2}$, and $\frac{\pi}{3}$ by hand, then $\frac{\pi}{4}$, $\frac{\pi}{5}$, and $\frac{\pi}{6}$ with a lot more scratch paper, I saw a pattern in the angles $\frac{\pi}{m}$ and wrote it down as a conjecture. Here it is, exactly as I wrote it.
For every positive integer $m$,
$$\sum_{n=1}^{\infty} \frac{(-1)^{\frac{n}{m} - 1}}{n} = \ln\left(1 - (-1)^{\frac{1}{m}}\right).$$This is the one place I wrote things with complex numbers. Here $(-1)^{\frac{1}{m}}$ means $\cos\frac{\pi}{m} + i \sin\frac{\pi}{m}$: the point $\frac{1}{m}$ of the way around the upper half of the unit circle. Its $n$-th power turns $n$ times as far, so $(-1)^{\frac{n}{m} - 1} = -\left(\cos\frac{n\pi}{m} + i \sin\frac{n\pi}{m}\right)$. Reading $a + bi$ as the vector $\begin{bmatrix} a \\ b \end{bmatrix}$, the left side is $-\sum \frac{1}{n} R\left(\frac{\pi}{m}\right)^{n} \begin{bmatrix} 1 \\ 0 \end{bmatrix}$: the harmonic Bender at $\theta = \frac{\pi}{m}$, turned one more step and flipped. So the conjecture is a closed form for the special values. And it's true. In fact there's a closed form for the harmonic Bender at every angle, and it only needs Calc 2.
For $0 < \theta < 2\pi$,
$$\sum_{n=1}^{\infty} \frac{\cos(n\theta)}{n} = -\ln\left(2\sin\tfrac{\theta}{2}\right) \qquad \text{and} \qquad \sum_{n=1}^{\infty} \frac{\sin(n\theta)}{n} = \frac{\pi - \theta}{2},$$and therefore
$$B_h(\theta) = R(-\theta) \begin{bmatrix} -\ln\left(2\sin\frac{\theta}{2}\right) \\[4pt] \frac{\pi - \theta}{2} \end{bmatrix}.$$Finite sums. For $t$ not a multiple of $2\pi$, the same telescoping as in the lemma on bounded sums gives
$$\frac{1}{2} + \sum_{k=1}^{N} \cos(kt) = \frac{\sin\left(\left(N + \frac{1}{2}\right)t\right)}{2\sin\frac{t}{2}}, \qquad \sum_{k=1}^{N} \sin(kt) = \frac{\cos\frac{t}{2} - \cos\left(\left(N + \frac{1}{2}\right)t\right)}{2\sin\frac{t}{2}}.$$Integrate. Fix $0 < \theta \leq \pi$ and integrate both identities over $[\theta, \pi]$, where $\sin\frac{t}{2} \geq \sin\frac{\theta}{2} > 0$. Write $g(t) = \frac{1}{2\sin(t/2)}$ and $\lambda = N + \frac{1}{2}$. Since $\sin(k\pi) = 0$ and $\cos(k\pi) = (-1)^k$,
$$\frac{\pi - \theta}{2} - \sum_{k=1}^{N} \frac{\sin(k\theta)}{k} = \int_\theta^\pi g(t) \sin(\lambda t)\, dt,$$ $$\sum_{k=1}^{N} \frac{\cos(k\theta)}{k} - \sum_{k=1}^{N} \frac{(-1)^k}{k} = \int_\theta^\pi \frac{\cos\frac{t}{2}}{2\sin\frac{t}{2}}\, dt - \int_\theta^\pi g(t) \cos(\lambda t)\, dt = -\ln\left(\sin\tfrac{\theta}{2}\right) - \int_\theta^\pi g(t) \cos(\lambda t)\, dt.$$The leftover integrals vanish. On $[\theta, \pi]$, $g$ and $g'$ are continuous, hence bounded. Integrating by parts,
$$\int_\theta^\pi g(t) \sin(\lambda t)\, dt = \left[-\frac{g(t)\cos(\lambda t)}{\lambda}\right]_\theta^\pi + \frac{1}{\lambda}\int_\theta^\pi g'(t) \cos(\lambda t)\, dt,$$and every piece on the right is a bounded quantity divided by $\lambda$, so it goes to $0$ as $N \to \infty$. The cosine integral is the same.
Let $N \to \infty$. The first line gives $\sum \frac{\sin(k\theta)}{k} = \frac{\pi - \theta}{2}$. In the second, $\sum \frac{(-1)^k}{k} = -\ln 2$ (the alternating harmonic series), so $\sum \frac{\cos(k\theta)}{k} = -\ln 2 - \ln\left(\sin\frac{\theta}{2}\right) = -\ln\left(2\sin\frac{\theta}{2}\right)$.
The other half. For $\pi < \theta < 2\pi$, put $\theta' = 2\pi - \theta \in (0, \pi)$. Then $\cos(k\theta) = \cos(k\theta')$, $\sin(k\theta) = -\sin(k\theta')$, and $\sin\frac{\theta}{2} = \sin\frac{\theta'}{2}$, so the formulas for $\theta'$ turn into the same formulas for $\theta$ (note $-\frac{\pi - \theta'}{2} = \frac{\pi - \theta}{2}$).
The Bender. Pulling the linear map $R(-\theta)$ out of the series (it comes out of every partial sum, so out of the limit), $B_h(\theta) = \sum \frac{1}{n} R(\theta)^{n-1} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = R(-\theta) \sum \frac{1}{n} R(\theta)^{n} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = R(-\theta) \begin{bmatrix} \sum \frac{\cos(n\theta)}{n} \\ \sum \frac{\sin(n\theta)}{n} \end{bmatrix}$. $\blacksquare$
The special values drop right out. At $\theta = \pi$: $R(-\pi)$ flips $\left(-\ln 2, 0\right)$ to $(\ln 2, 0)$. At $\theta = \frac{\pi}{2}$: $2\sin\frac{\pi}{4} = \sqrt{2}$, and $R\left(-\frac{\pi}{2}\right)$ turns $\left(-\ln\sqrt{2}, \frac{\pi}{4}\right)$ into $\left(\frac{\pi}{4}, \ln\sqrt{2}\right)$. (The $\sqrt{2}$ in $\ln\sqrt{2}$ really is the natural way to write it.) At $\theta = \frac{\pi}{3}$: $2\sin\frac{\pi}{6} = 1$, the logarithm disappears entirely, and $B_h\left(\frac{\pi}{3}\right) = R\left(-\frac{\pi}{3}\right)\left(0, \frac{\pi}{3}\right) = \left(\frac{\pi}{2\sqrt{3}}, \frac{\pi}{6}\right)$.
Blake's Conjecture is true, for every positive integer $m$, with $\ln$ the principal logarithm.
Let $\theta = \frac{\pi}{m} \in (0, \pi]$. By the theorem, the left side is
$$-\sum_{n=1}^{\infty} \frac{\cos(n\theta) + i\sin(n\theta)}{n} = \ln\left(2\sin\tfrac{\theta}{2}\right) - i\,\frac{\pi - \theta}{2}.$$On the right, $1 - (-1)^{\frac{1}{m}} = 1 - \cos\theta - i\sin\theta = 2\sin\frac{\theta}{2}\left(\sin\frac{\theta}{2} - i\cos\frac{\theta}{2}\right) = 2\sin\frac{\theta}{2}\left(\cos\frac{\theta - \pi}{2} + i\sin\frac{\theta - \pi}{2}\right)$. That's the number with length $2\sin\frac{\theta}{2}$ and angle $\frac{\theta - \pi}{2} \in \left(-\frac{\pi}{2}, 0\right]$, so its principal logarithm is $\ln\left(2\sin\frac{\theta}{2}\right) + i\,\frac{\theta - \pi}{2}$. The two sides agree. $\blacksquare$
Notice the proof never used that $\theta$ was $\frac{\pi}{m}$: the same identity holds with $(-1)^{\frac{1}{m}}$ replaced by $\cos\theta + i\sin\theta$ for any $0 < \theta < 2\pi$. Here are the values from my notes, now in exact form.
| $m$ | $B_h\left(\frac{\pi}{m}\right)$, exactly | approximately |
|---|---|---|
| $1$ | $(\ln 2,\ 0)$ | $(0.6931,\ 0)$ |
| $2$ | $\left(\frac{\pi}{4},\ \ln\sqrt{2}\right)$ | $(0.7854,\ 0.3466)$ |
| $3$ | $\left(\frac{\pi}{2\sqrt{3}},\ \frac{\pi}{6}\right)$ | $(0.9069,\ 0.5236)$ |
| $4$ | $\frac{\sqrt{2}}{2}\left(\frac{3\pi}{8} - \frac{1}{2}\ln(2 - \sqrt{2}),\ \ \frac{3\pi}{8} + \frac{1}{2}\ln(2 - \sqrt{2})\right)$ | $(1.0221,\ 0.6440)$ |
| $5$ | $\left(\frac{\varphi}{2}\ln\varphi + \frac{2\pi}{5}s,\ \ -s\ln\varphi + \frac{\pi\varphi}{5}\right)$, with $\varphi = \frac{1 + \sqrt{5}}{2}$, $s = \sin\frac{\pi}{5}$ | $(1.1279,\ 0.7338)$ |
| $6$ | $\left(\frac{5\pi}{24} - \frac{\sqrt{3}}{4}\ln(2 - \sqrt{3}),\ \ \frac{5\sqrt{3}\,\pi}{24} + \frac{1}{4}\ln(2 - \sqrt{3})\right)$ | $(1.2248,\ 0.8044)$ |
The golden ratio sneaking in at $m = 5$ comes from $2\sin\frac{\pi}{10} = \frac{1}{\varphi}$, so $-\ln\left(2\sin\frac{\pi}{10}\right) = \ln\varphi$. Here is the whole image of the harmonic Bender, with the conjecture's points lighting up one at a time.
The closed form also explains the tongue. As $\theta \to 0^+$, the $x$-coordinate $-\ln\left(2\sin\frac{\theta}{2}\right)\cos\theta + \frac{\pi - \theta}{2}\sin\theta$ grows without bound, while the $y$-coordinate $\ln\left(2\sin\frac{\theta}{2}\right)\sin\theta + \frac{\pi - \theta}{2}\cos\theta$ goes to $\frac{\pi}{2}$, since $\sin\theta \ln\left(2\sin\frac{\theta}{2}\right) \to 0$. By the mirror symmetry, the other side approaches $-\frac{\pi}{2}$.
Multiplying by $\cos\theta + i\sin\theta$ turns the complex plane exactly the way $R(\theta)$ turns $\mathbb{R}^2$. So with $z = \cos\theta + i\sin\theta$, the Bender of a sequence is the power series $\sum a_n z^{n-1}$, evaluated on the unit circle. Geometric sequences give $\frac{1}{1 - rz}$, and the circle images are the classic fact that $w \mapsto \frac{1}{1 - w}$ sends circles to circles. The harmonic sequence gives $-\frac{\ln(1 - z)}{z}$, the logarithm's power series, and evaluating it on the unit circle is exactly what Abel's theorem allows once you know the series converges there.
Here's the honest ending. Blake's Conjecture turns out to be a classical identity. The series $\sum \frac{\sin(n\theta)}{n} = \frac{\pi - \theta}{2}$ goes back to Euler, the logarithm's power series to Mercator in 1668, and Abel's theorem (1826) is what lets you plug a point of the unit circle into it. I found it independently, as an undergrad, using nothing but Calculus 2 series and rotation matrices, and I didn't know any of that history until much later.
What the Bender adds is the route. Bending a sequence by an angle turns old facts into pictures you can actually watch: a geometric series traces a circle, because peeling off one arrow leaves the whole chain turned and shrunk; the circle test turns "it looks round" into a proof; and the harmonic series traces a tongue that runs off to infinity, creeping up on two horizontal lines, passing through $\ln 2$, Leibniz's $\frac{\pi}{4}$, and the golden ratio along the way.
And some questions are still open, at least in this framing:
There's another question I wonder about, and it connects the Bender to dynamical systems. What if the turn was random? What would the distribution of curves look like? Does that question even make sense?
It does, but first you have to decide where the randomness goes. Here's the way I see it now. The direction of the $n$-th arrow is the running total of the turns before it, $\Phi_n = \theta_1 + \theta_2 + \cdots + \theta_{n-1}$, so the Bender is
$$B = \sum_{n=1}^{\infty} a_n \begin{bmatrix} \cos\Phi_n \\ \sin\Phi_n \end{bmatrix}:$$a weighted sum of one fixed function of the angle, read off along an orbit. Dynamical systems people call a sum like that a Birkhoff sum. With a fixed turn, the orbit is just a rotation of the circle, $\varphi \mapsto \varphi + \theta$, about the simplest dynamical system there is. Asking for a random turn means swapping that rotation for a random walk on the circle, or for something chaotic. I think there are three versions worth keeping apart.
Version 1: every turn completely random. If each turn is independent and uniform on the circle, then each arrow points in an independent, uniformly random direction, and the turning structure disappears. What's left is a sum of independent random vectors, and that's classical.
If the arrows point in independent, uniformly random directions, then $\mathbb{E}\,\lVert B_N \rVert^2 = \sum_{n=1}^{N} a_n^2$ for the partial sums $B_N$. If $\sum a_n^2 < \infty$, the random Bender converges with probability $1$.
Write $u_n$ for the direction of the $n$-th arrow. Expanding, $\lVert B_N \rVert^2 = \sum_{n} a_n^2 \lVert u_n \rVert^2 + \sum_{n \neq k} a_n a_k\, u_n \cdot u_k$. Each $\lVert u_n \rVert^2 = 1$, and for $n \neq k$, $u_n \cdot u_k = \cos(\Phi_n - \Phi_k)$ averages to $0$ because the angle between two independent uniform directions is uniform. That gives $\mathbb{E}\,\lVert B_N \rVert^2 = \sum_{n \leq N} a_n^2$.
Each new arrow has average $0$ given everything before it, so the partial sums $B_N$ form a martingale, and the computation above shows $\mathbb{E}\,\lVert B_N \rVert^2$ stays bounded when $\sum a_n^2 < \infty$. A martingale bounded in this mean-square sense converges with probability $1$ (the martingale convergence theorem). $\blacksquare$
The harmonic punchline: $\sum \frac{1}{n^2} = \frac{\pi^2}{6} < \infty$, so the harmonic Bender with random turns converges with probability $1$, with $\mathbb{E}\,\lVert B \rVert^2 = \frac{\pi^2}{6}$, even though the fixed-turn version blows up at $\theta = 0$. Randomness tames it.
Version 2: each turn a coin flip between $0$ and $\pi$. Then every arrow points left or right at random, and the Bender is $\sum \pm a_n$ on the $x$-axis. For the harmonic sequence, that's the random harmonic series $\sum \pm \frac{1}{n}$, which is a known object: it converges with probability $1$, and its distribution has a smooth density with some famous oddities, like a value at $2$ that comes astonishingly close to $\frac{1}{8}$ without being equal to it (Schmuland, 2003). The Bender turns this into a whole family: let each turn be $\theta$ or $-\theta$ at random, and the endpoints spread out into the plane.
Version 3: a fixed turn with a little noise. This is the closest to what I first had in mind. Make each turn $\theta$ plus a small random wobble. Now each angle gives a random cloud of endpoints instead of a single point, so "the distribution of curves" becomes concrete: as the noise grows, the sharp image (the circle, or the harmonic tongue) should smear into a fuzzy band, and eventually into the round blob of Version 1. How thick is the band for a given amount of noise? And does any noise at all tame $\theta = 0$ for the harmonic sequence? I suspect it does: with a Gaussian wobble of size $\sigma$, the correlation between arrows $n$ and $k$ should decay like $e^{-\sigma^2 |n - k|/2}$, which ought to keep $\mathbb{E}\,\lVert B \rVert^2$ finite. That's a guess, not a theorem yet.
Bonus: chaos instead of chance. Instead of a random walk, let the directions follow the doubling map, $\Phi_n = 2^{n-1}\varphi$. Then the Bender becomes a function of the starting angle $\varphi$, and for $a_n = \left(\frac{1}{2}\right)^{n-1}$ its coordinates are $\sum \left(\frac{1}{2}\right)^{n-1} \cos\left(2^{n-1}\varphi\right)$ and the matching sine sum, which are Weierstrass functions: continuous everywhere, differentiable nowhere (Hardy's theorem). The smooth image curves turn into rough, fractal-looking ones.
So the question makes sense. It's really three questions, and Version 3 is where I'd start.