Prove $f(x)=\sum_{k=1}^\infty \frac{1}{k!\sqrt{k}}x^k $ defines a continuous function on $\mathbb{R}$.

Question

Prove $$f(x)=\sum_{k=1}^\infty \frac{1}{k!\sqrt{k}}x^k$$ defines a continuous function on $\mathbb{R}$.

I think we can show that if $\sum_{k=1}^\infty \frac{1}{k!\sqrt{k}}x^k $ is uniformly convergent $\forall x\in\mathbb{R}$, then it defines a continuous function on $\mathbb{R}$.

We could say that $$\sum_{k=1}^\infty \frac{1}{k!\sqrt{k}}x^k \le \sum_{k=1}^\infty \frac{1}{k!}x^k $$

And show that $\sum_{k=1}^\infty \frac{1}{k!}x^k$ converges pointwise by the ratio test, so $f(x)$ must converge pointwise. But how would I show uniform convergence? If I show uniform convergence, there's a theorem that lets me say that uniform limits of continuous functions are continuous (thus $f(x)$ is continuous.)

Thanks in advance.

Answer 1

Your series does not converge uniformly on $\Bbb R$.

But, given $R>0$, it does converge uniformly on $[−R,R]$, by the result mentioned in one of your previous comments above (which is essentially the M-test).

Thus for any $R>0$, $f$ is continuous on $[−R,R]$ (for reasons mentioned at the end of your post). This implies $f$ is continuous at each $x\in\Bbb R$; and thus, $f$ is continuous on $\Bbb R$.

Condensed version: You need to show $f$ is continuous at each fixed $x\in \Bbb R$. This will follow if you can show the convergence is uniform on any set of the form $[-R,R]$, $R>0$.

Answer 2

For a shorter argument (requires more knowledge though), we may say that we're dealing with a power series with infinite radius of convergence (use ratio test to prove this).

But a power series converges uniformly over any compact set included in the open disk of convergence.

It is therefore continous on any compact subset of $\mathbb R$, hence continuous at any point in $\mathbb R$.