Seeking a Geometric Proof of a Generalized Alternating Series' Convergence

Question

Let $z \in \mathbb{C} \backslash \lbrace 1 \rbrace$ with $|z| = 1$. We consider the following infinite series, which necessarily converges: $$S(z) := \sum_{n = 1}^{\infty}\frac{z^n}{n}$$

Note that $S(-1)$ is the alternating harmonic series.

A straightforward application of the Dirichlet Convergence Test proves any such series converges, but I feel this is a bit like killing a fly with a sledgehammer. (I realize some of you might not think this test is a sledgehammer; I wonder also whether this series is a fly.) In any event, I'm wondering whether there is a way to prove convergence using only a simple geometric argument (with some basic analysis).

For example, we can think of $S(i)$ as taking steps in the plane of length $1/n$, but turning ninety degrees after each one. Then the partial sums correspond to a nested sequence of squares, where the area of the squares is clearly converging to $0$. Thus, an argument using the Nested Interval Property (or really its corresponding $2D$ version) indicates that the series converges.

More generally, I'd think that because we are taking steps of size decreasing to $0$ and rotating by the same amount after each step, there should be a general geometric argument for why $S(z)$ will converge. Ideally, I'd like to have a proof that could be made accessible to early Calculus students, even if not every step is presented in fully rigorous form.

For clarity's sake, I will directly state my question: How does one prove $S(z)$ converges using a simple geometric argument that relies at most on basic analysis (e.g., makes no appeal to stronger theorems from Complex Analysis)?

Answer 1

Here is what I think you are looking for:

First, note that if you take steps of fixed length $\ell$, and keep rotating by an angle of $\theta\neq 0$, then you will stay inside a circle of radius $r=\frac{\ell}{2}\cos(\theta/2)^{-1}$. In fact, the steps will all land on the circle, and you can calculate its center as the point of distance $r$ from one of the steps, at an angle which bisects the angle $\theta$. Now if at a certain step you decide to change the length of your steps to $\ell'<\ell$, the new circle will have radius $r'=\frac{\ell'}{2}\cos(\theta/2)^{-1}$, with center the point of distance $r'$ away from the current step in the same direction as the old center. From this description it's clear that the new circle is contained in the old circle. Now you can apply the nested interval property.

Answer 2

This proof is not really much more than a rewriting of Dirichlet's test, but here goes:

$$(z-1) \sum_{n=1}^N \frac{z^n}{n} = \sum_{n=2}^{N+1} \frac{z^{n}}{n-1} - \sum_{n=1}^N \frac{z^n}{n} = \frac{z^{N+1}}{N} - \frac{z}{1}+ \sum_{n=2}^N \frac{z^n}{n(n-1)}$$

which converges as $N \to \infty$.

Answer 3

Here's an idea. Group the series into blocks

$$\sum_{n=dk}^{d(k+1) - 1} \frac{z^n}{n}$$

where $d$ is fixed and large enough that the complex numbers $1, z, z^2, ... z^{d-1}$ are approximately uniformly distributed across the unit circle. The terms in each block should then be approximately uniformly distributed in phase across the unit circle, and in particular each term should be pairable with a term approximately the negative of it, the two of them canceling to order $O \left( \frac{1}{n^2} \right)$.

But making this precise seems to me to require more effort than proving that Dirichlet's convergence test works.

Answer 4

Think of an alternating series

$$\sum_{n\geq 0}(-1)^n a_n,\;\;a_n>a_{n+1}>0 $$

as describing the motion of a hesitating person, who starts at $a_0$, and goes alternatively, one step backward, one step forward. If we denote by $S_n$ its location after $n$-steps

$$ S_n=\sum_{k=0}^n (-1)^k a_k, $$

then we observe that

$$ S_0>S_1>0 $$

and we deduce inductively

$$ S_{2n}> S_{2n+2}>S_{2n+1} >0 $$

Thus during the travel, the person is always on the right-hand side of the origin, and every two steps he gets closer to the origin. Mathematically, this means that the subsequence $S_{2n}$ is decreasing and positive thus it has a limit. The odd subsequence $S_{2n+1}$ converges to the same limit since

$$ S_{2n}-S_{2n+1}=a_{2n+1}\to 0. $$

We can visualize this process by considering the continuous function $\newcommand{\bR}{\mathbb{R}}$ $S:[0,\infty)\to \bR$ which is linear on each of the intervals $[n,n+1]$, $n$ nonegative integer, and such that $S(n)= S_n$. It can be visualized as a zig-zag, down-up-down-up, that stays above the $x$-axis, while the peaks are getting shorter and shorter.