# How to show that $\sum\limits_{n=1}^\infty \exp(i\,nz)$ converges?

How do I show that $$f(z) = \exp(i\, z) + \exp(i\, 2z) + \ldots + \exp(i\, nz) + \ldots$$ converges? Problem is taken from a Yahoo! Answers question: "Find the infinite sum of sin(n)/n?".

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In the Y! answer, it seem that user "rodolfo riverol" is using the expansion $\frac{x}{1-x} = x + x^2 + \ldots$ with $x = \exp(iz)$ and $x^k = \exp(ikz).$ – user2468 Aug 25 '12 at 21:51
does that work for complex variables? don't we have a different sets of test for complex series? – hasExams Aug 25 '12 at 21:52
You could take a look at: en.wikipedia.org/wiki/Geometric_series#Formula – Thomas Aug 25 '12 at 21:53

For the series to converge, you have the following condition, $$|{\rm e}^{iz}| < 1 \,.$$

Assuming $z=x+iy$, we have,

$$\left|{\rm e}^{i(x+iy)} \right| = \left|{\rm e}^{ix-y)} \right| = {\rm e}^{-y} < 1 \Rightarrow -y < 0 \Rightarrow y > 0 \,.$$

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It looks like a geometric series to me. Recall that a geometric series is of the form

$S = a + ar + ar^2 + ar^3 + \cdots \, .$

In this case, the first term $a = e^{iz}$ while the common ratio $r = e^{iz}.$

$S = \frac{a}{1-r} = \frac{e^{iz}}{1-e^{iz}} \, .$

This function defines the analytic continuation of $f$ provided $e^{iz} \neq 1$.

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Nop. There's a very basic condition the constant ratio of an infinite geometric sequence must fulfill in order to have that its (infinite) series converges... – DonAntonio Aug 25 '12 at 23:03
I assume you mean $|r| < 1$. This is a sufficient condition for $S$ to converge, just like $|z| < 1$ is a sufficient condition for $1 + z + z^2 + \cdots$ to converge. However, $1 + z + z^2 + \cdots$ has an analytic continuation, namely $(1-z)^{-1},$ for all $z \in \overline{\mathbb{C}}\backslash\{1\}$. Just like the OP's function has an analytic continuation for all $e^{iz} \neq 1.$ – Fly by Night Aug 26 '12 at 1:12
Ok @Fly, I'm confused: what are you calling $\,"f"\,$ to in your answer, then? The condition $\,|r|<1\,$ is a sufficient and necessary condition for convergence of an infinite geometric series. I think you may mean that $\,f\,$ is the infinite series in the OP...? I think this isn't clear enough. – DonAntonio Aug 26 '12 at 1:50
@DonAntonio: I think he means that $f$ as a function of $z$ is well defined by the series for $z$ such that $\lvert e^{iz}\rvert<1$, and for all other $z$ except $2k\pi$, it can be meaningfully extended (even if the original series does not converge). – tomasz Aug 26 '12 at 1:53
@tomasz, I think it is that. Thanks. – DonAntonio Aug 26 '12 at 2:18

The general proof for the formula for the geometric series given here on Wikipedia works for complex numbers as well. If you look at real numbers: $1 + r + r^2 + ...$, then we need $\lvert r \lvert < 1$. If $r$ is a complex number, then you still need $\lvert r \lvert < 1$, where now the $\lvert \cdot\lvert$ is the complex norm.

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The radius of convergence of this series is $1$, due to the pole at $1$ of the geometric series. Thus, for any $z$ such that: $$|\exp(iz)| < 1$$ the series converges. This is true for every $y = \Im(z) > 0$, since for these values, you have a factor $e^{i(iy)}=e^{-y} < 1$.

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Whenever $z$ is real, $|\exp(iz)|=1$ exactly, not less than. Instead of being $\ne k\pi$, we need $z$ to be somewhere in the (open) upper half-plane. – Henning Makholm Aug 25 '12 at 22:08
@HenningMakholm - fixed. Must have been sleeping. – nbubis Aug 25 '12 at 22:31
Okay. I took the liberty to edit in a further correction. – Henning Makholm Aug 25 '12 at 22:36