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Consider the series for $x \in \mathbb{R}$

$$ \sum_{n=1}^\infty \frac{1}{\sqrt{n}}\left(e^{-\frac{x^2}{n}}-1 \right) $$ Then I have to prove that the series converges pointwise on $\mathbb{R}$. To prove this is all I need is to let $x \in \mathbb{R}$ and prove that the limit $$ \lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}}\left(e^{-\frac{x^2}{n}}-1 \right) $$ exists? It is obvious that $\lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}} \rightarrow 0$ and $\lim_{n \rightarrow \infty} e^{\frac{-x^2}{n}} -1 \rightarrow 0$ for $x \in \mathbb{R}$ and thus the product of the limits is also equal to $0$. Thus the series must converge pointwise. Is this ok? Or do I need anything else? Thanks for your help.

Mathias
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  • $a_n\to 0$ doesn't imply that $\sum_{n=1}^\infty a_n$ converges. Can you compare the decay of $\frac{1}{\sqrt{n}}\left(e^{-\frac{x^2}{n}}-1 \right)$ and $\frac1{n^c}$ for say $c=4/3$ ? – reuns May 16 '20 at 15:17
  • Would I have to use asymptotic equivalents as Bernard did? I am not sure we have even heard that in our course. Thus I may not be able to use it. – Mathias May 16 '20 at 15:29
  • Would I be able to use Weiterstrass M-test to prove this as uniform convergence also implies pointwise convergences? – Mathias May 16 '20 at 15:32

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That is not enough. From $\frac1{\sqrt n}\to 0$, you cannot deduce that the series $\sum\limits_{n\ge 1}\frac1{\sqrt n}\cdot\frac1{\sqrt n}=\sum\limits_{n\ge 1}\frac1n$ converges. You can obtain the convergence in the case of your series by considering asymptotic equivalents:

We have $e^{-\frac{x^2}{n}}-1\sim_\infty -\frac{x^2}n$, so $$\frac{1}{\sqrt{n}}\Bigl(\mathrm e^{-\tfrac{x^2}{n}}-1 \Bigr)\sim_\infty\frac{1}{\sqrt{n}}\biggl(-\frac{x^2}{n}\biggr)=-\frac{x^2}{n^{\tfrac 32}}, $$ which is the general term of a convergent $p$-series.

Bernard
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