Pointwise or uniform convergence of $\sum_{n = 0}^{\infty} \frac{(ix)^n}{n!}$

Question

I am a bit curious about whether the convergence of $\sum_{n = 0}^{\infty} \frac{(ix)^n}{n!}$ converges uniformly or pointwise to $e^{ix}$. I have thought about the convergence of both $\sum_{n = 0}^{\infty} \frac{(-1)^nx^{2n + 1}}{(2n + 1)!}$ and $\sum_{n = 0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}$ which both converge pointwise to $\sin(x)$ and $\cos(x)$. But then I thought that since each $\frac{(-1)^nx^{2n + 1}}{(2n + 1)!}$ and $\frac{(-1)^nx^{2n}}{(2n)!}$ are continuous and of course, $\sin(x)$ and $\cos(x)$ are both continuous, so that should mean that $\sum_{n = 0}^{\infty} \frac{(-1)^nx^{2n + 1}}{(2n + 1)!}$ and $\sum_{n = 0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}$ should converge uniformly. So based on these observations, I was wondering how $\sum_{n = 0}^{\infty} \frac{(ix)^n}{n!}$ would behave in terms of convergence, whether it'd be uniform or not, since it converges pointwise to $\cos(x) + i\sin(x)$. Am I missing something here? I would receive some advice on this.

Answer 1

As is typically the case with power series, the convergence is uniform on compact sets. If we consider $x$ as a real variable only, then this would translate to uniform convergence on bounded closed intervals, for example.

There is a general theorem for the uniform convergence of power series. A simple statement and proof of such a theorem (for real coefficient series, although adapting to complex coefficients is not very different) is in Abbott's Understanding Analysis, Theorem 6.5.4 (called Abel's Theorem):

Let $g(x) = \sum_{n=0}^\infty a_nx^n$ be a power series that converges at the point $x = R>0$. Then the series converges uniformly on the interval $[0,R]$. Similarly, if the series converges at the point $x = -R < 0$, then there is uniform convergence on $[-R,0]$.

The idea of the proof is that if $x \in [0,R]$, then $|x/R| < 1$ and so you can replace the term $a_n x^n$ with $a_nR^n\cdot(x/R)^n$. By hypothesis, the series $\sum_{n=0}^\infty a_nR^n$ converges, and so multiplying each term by the shrinking value $(x/R)^n$ does not change this rate of convergence.

However, I should remark that even though you get uniform convergence on compact sets, you do NOT get such on all of $\mathbb{R}$. In this case this is because $e^{ix}$ is a bounded function, yet every partial sum of the series is a polynomial and hence unbounded. So you only get pointwise convergence on all of $\mathbb{R}$.

Answer 2

Non-uniform Convergence for $x \in \mathbb{R}$

Note that a series $\sum_{n=0}^\infty f_n(ix)$ converges uniformly for all $x \in \mathbb{R}$ if and only if the sequence of partial sums satisfies the uniform Cauchy criterion:

For every $\epsilon>0$ there exists $N(\epsilon)\in \mathbb{N}$ such that if $m \geqslant n \geqslant N(\epsilon)$, then for every $x \in \mathbb{R}$ we have

$$\left|\sum_{k = n}^m f_k(ix) \right|< \epsilon $$

Taking $m = n$ we get as a necessary condition for uniform convergence of the series, that $f_n(ix) \to 0$ uniformly for all $x \in \mathbb{R}$ which is equivalent to $\sup_{x \in \mathbb{R}}|f_n(ix)| \to 0$ as $n \to \infty$.

However, with $f_n(ix) = \frac{(ix)^n}{n!}$ we have as $n \to \infty$

$$\sup_{x \in \mathbb{R}}|f_n(ix)| = \sup_{x \in \mathbb{R}}\frac{|(ix)^n|}{n!}= \sup_{x \in \mathbb{R}}\frac{|x|^n}{n!}= +\infty \not\to 0$$

Hence, the series is not uniformly convergent on $\mathbb{R}$.