Check the uniform convergence of $\sum\limits_{n = 1}^{\infty} \frac{n^2}{x} \exp(\frac{-n^2}{x})$

Question

$$\sum\limits_{n = 1}^{\infty} \frac{n^2}{x} \exp{\frac{-n^2}{x}}$$ where $ 0 < x < \infty$

While $$\lim\limits_{n\rightarrow\infty} (a_n)^{1/n} = 0$$ The sum would converge. But how to check uniform convergence?

Answer 1

The series does not converge uniformly on $(0,\infty)$. I will prove it showing that Cauchy's uniform criterion is not satisfied. For any $n\in\mathbb{N}$ $$\begin{align} \sup_{x>0}\sum_{k=n}^{2n}\frac{n^2}{x}e^{-n^2/x}&\ge\sup_{x>0}\,(n+1)\,\frac{n^2}{x}e^{-4n^2/x}\\ &\ge(n+1)\,\sup_{z>0}z\,e^{-4z}\\ &\ge\frac{n+1}{4\,e}. \end{align}$$ Thr convergence is uniform on any interval $[a,\infty)$, $a>0$.

Answer 2

Let $A$ and $B$ verifying $B>A>0$, consider $x \in [A,B]$. Using $$e^{-n^2/x}\leq e^{-n^2/B}$$ implying $$\sum_{n=1}^{\infty}n^2e^{-n^2/x}\leq \sum_{n=1}^{\infty}n^2e^{-n^2/B} $$ gives normal convergence of your series on each compact set $[A,B]$ and thus you have uniform convergence on any $[A,+\infty).$