# Convergence of a power series in the complex plane — does this argument work?

I'm interested to know for which $z \in \mathbb{C}$ the power series $\sum_{n=0}^{\infty}\frac{\cos(nz)}{n!}$ converges and what its value is when it does.

I've come up with the following argument, according to which the power series converges for all $z \in \mathbb{C}$, but the wording of the question has me a bit skeptical as to whether I'm correct.

$$\sum_{n=0}^{\infty}\frac{\cos(nz)}{n!}=\sum_{n=0}^{\infty}\frac{e^{inz}+e^{-inz}}{2n!}=\frac{1}{2}\sum_{n=0}^{\infty}\frac{(e^{iz})^n}{n!}+\frac{1}{2}\sum_{n=0}^{\infty}\frac{(e^{-iz})^n}{n!}=\frac{1}{2}e^{e^{iz}}+\frac{1}{2}e^{e^{-iz}}$$

Sorry that this is all on the same line. Do you buy it?

Thanks.

Edit:

In response to Qiaochu, I've put together the following inequality:

\begin{align} |cos(nz)| &= |\frac{e^{inz}+e^{-inz}}{2}|\\ &=\frac{1}{2}|e^{inz}+e^{-inz}|\\ &\leq \frac{1}{2}|e^{inz}|+\frac{1}{2}|e^{-inz}|\\ &=\frac{1}{2}|e^{-nIm(z)}e^{inRe(z)}|+\frac{1}{2}|e^{nIm(z)}e^{-inRe(z)}|\\ &=\frac{1}{2}|\frac{1}{e^{nIm(z)}}|+\frac{1}{2}|e^{nIm(z)}|\\ &=\frac{1}{2}(\frac{1}{e^{nIm(z)}}+e^{nIm(z)})\\ \end{align}

For $z$ with $Im(z)=0$, this gives us that that $|\frac{cos(nz)}{n!}|\leq \frac{1}{n!}$ for each $n$. Since $\sum_{n=0}^{\infty}1/n!=e$, the series converges (absolutely). Of course, this would have been obvious without having worked that all out.

How do I use the bound in the general case? Do I just say that the factorial denominator will grow much faster than this exponential numerator, and hence the series will converge?

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You can't manipulate the series until you know that it converges. Try bounding $|\cos (nz)|$. –  Qiaochu Yuan Jan 17 '12 at 22:37
It's fine. The series $\sum_n (e^{iz})^n/n!$ and $\sum_n (e^{-iz})^n/n!$ converge, therefore so does their term-by-term sum... –  Robert Israel Jan 17 '12 at 22:46
Looks good. Power series for $e^w$ converges for all complex $w$. Might be worth mentioning explicitly. I wouldn't call your original series a power series, or even a sum of two of them. –  André Nicolas Jan 17 '12 at 22:54
Thanks all. I think I'll type up the bounding of $|cos(nz)|$ once I've worked it out. –  user18297 Jan 17 '12 at 22:56
@Robert: okay, but then the proof as written works slightly backwards. The first equality isn't justified until you know that the leftmost expression is well-defined. –  Qiaochu Yuan Jan 18 '12 at 0:03