# uniform convergence of series of functions

Prove that $f(x)=\displaystyle{\sum_{n=1}^{\infty}} \dfrac{e^x \sin (n^2x)}{n^2}$ is convergent for every $x \in \mathbb{R}$ and that its sum $f(x)$ is a continuous function on $\mathbb{R}$.

This is my tentative to solve the problem: $f(x)= e^x \sum {\frac{\sin (n^2x)}{n^2}}$. So, to prove that $f(x)$ is convergent, I only need to prove that $\sum {\frac{\sin(n^2x)}{n^2}}$ is convergent. since absolute value of $\frac {\sin (n^2x)}{n^2} \le \frac{1}{n^2}$ because absolute value of $\sin (n^2x)$ is $\le 1$ and the series: $\sum {\frac{1}{n^2}}$ is convergent, then by the $M$-test the series $\sum {\frac{\sin (n^2x)}{n^2}}$ is uniformly convergent and thus $f(x)$ is continuous on $\mathbb{R}$.

Please let me know whether my solution is true? Also, do I have to distinguish the two cases where $x=0$ and $x \ne 0$? Do I have to prove that $f$ is continuous at $x=0$ separately?

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@Srivatsan I edited LaTeX only. I think, you should also notice the spelling "beccause". – gaurav Nov 14 '11 at 3:18

You're very close. Minor notes:

• No, you don't have to consider $x=0$ as a special case, because $|\sin(0)|\leq 1$ is still true.
• $f(x)= e^x \sum(\sin(n^2x)/n^2)$ is true, but note now that the series is different. You gave an argument for why the series here converges uniformly, but note that this does not imply that the initial series defining $f(x)$ converges uniformly (and it does not).
• Related to the last point, you technically never addressed why the original series converges for every $x$. This follows from the absolute convergence test, or from the work you already did and the fact that if $\sum a_n$ converges and $b\in \mathbb R$, then $\sum ba_n$ converges to $b\sum a_n$. (In other words, you could just make a little more explicit some of the work you already did. How much detail to include is largely a matter of taste.)
• Other details that may be mentioned to finish this off (perhaps intentionally omitted in your question because they are straightforward):
• For each $n$, $x\mapsto \sin(n^2x)/n^2$ is continuous, which is why you can conclude that the uniformly convergent series $\sum\sin(n^2 x)/n^2$ is continuous.
• $x\mapsto e^x$ is continuous.
• The product of 2 continuous functions is continuous.
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+1. "this does not imply that the initial series defining $f(x)$ converges uniformly (and it does not)" - So this gives an example where a sequence of continuous functions converges to a continuous limit, but the convergence is not uniform. In fact, this might even the hidden context of the question. – Srivatsan Nov 14 '11 at 3:15
Srivatsan makes a good point. Here at least there is uniform convergence on compact subsets which is also generally strong enough to conclude that the limit is continuous (and fortunately as the OP found there is a way to avoid even worrying about that). For examples where even uniform convergence on compact subsets fails but the limit is still continuous, take a sequence $f_n:[0,1]\to \mathbb R$ such that $f_n$ increases linearly from $0$ to $1$ on the interval $[0,1/n]$, decreases linearly from $1$ to $0$ on the intreval $[1/n,2/n]$, and is $0$ on $[2/n,1]$. Then $f_n\to 0$ nonuniformly. – Jonas Meyer Nov 14 '11 at 3:24
Ok. great. So this proves that the function f(x) is continuous because it is the product of two continuous functions. But, How does it follow that f(x) is convergent? and what type of convergence f is following? Is it pointwise convergence? Can anyone help please? – M.Krov Nov 14 '11 at 17:25
@Zi2018Alpha: I added a new bullet on convergence of the original series. The problem says "is convergent for every $x$," so yes it means pointwise convergence. That is, for each (fixed) $x$, the series $\displaystyle{\sum_{n=1}^{\infty}} \dfrac{e^x \sin (n^2x)}{n^2}$ converges. The absolute convergence test is one way to show this. It is also true in general that if $\sum a_n$ converges and $b\in\mathbb R$, then $\sum ba_n$ converges to $b\sum a_n$, so you don't need to directly analyze the original series (at least if you can justify my claim). – Jonas Meyer Nov 14 '11 at 18:17