Examine convergence and (almost) uniform convergence of $\sum_{n=1}^{+ \infty} \frac{n^2 x^2}{e^{n^2 |x|}}$ for $x \in \mathbb{R}$.

Question

How to examine convergence, almost uniform convergence and uniform convergence of series $\sum_{n=1}^{+ \infty} \frac{n^2 x^2}{e^{n^2 |x|}}$ for $x \in \mathbb{R}$?

Answer 1

You can intuitively get an idea of the result by analyzing the convergence for different values of $x$.

• If $x=0$ then the sum is equal to 0 (convenient)
• if $x\neq 0$ then $\frac {n^2x^2} {e^{n^2\vert x\vert}}=o(\frac 1 {n^2})$ for instance, so the series is convergent

So the function series is convergent.

Now since $\frac {n^2x^2} {e^{n^2\vert x\vert}}$ is very, very small for any $x$, let us show that it is unifornly convergent. Actually we are going to prove that there is normal convergence (https://en.wikipedia.org/wiki/Normal_convergence) which is even stronger (and simpler). The idea is to find $(a_n)_{n\in\mathbb N}$ such that $\forall n, \frac {n^2x^2} {e^{n^2\vert x\vert}}<a_n$.

Indeed, for any $n\in \mathbb N$, the function $x\longrightarrow \frac {n^2x^2} {e^{n^2\vert x\vert}}$ is continuous, pair and has limit 0 in both $\infty$ and $-\infty$. So it is bounded. Let us try and show that this bound is dependent of $n$. Since the function is pair, we restrict our analysis on $]0,\infty[$, where it is (conveniently) differentiable.

Let us look for the zero of the derivative :

$$(x\longrightarrow \frac {n^2x^2} {e^{n^2\vert x\vert}})'(x)=\frac {n^2x(2-2n^2x)}{e^{2n^2\vert x\vert}}$$

The only strictly positive root of this equation is... $\frac 1 {n^2}$

So $\forall n\in\mathbb N,\forall x\in\mathbb R,\frac {n^2x^2} {e^{n^2\vert x\vert}}\leqslant\frac {n^2(\frac 1 {n^2})^2} {e^{n^2\vert \frac 1 {n^2}\vert}}=\frac 1 {n^2e}$. We have normal convergence for $a_n=\frac 1 {n^2e}$.

So there is uniform convergence.