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I was asked to show whether the following series converges: $$\sum_{n=1}^{\infty}\frac{1}{{e^{n^2}}}$$

Here's my first attempt which I thought was pretty straightforward: $$e^{n^{2}}=(e^{n})^{n}\ge e^{n} \ ,\ \forall \ n \in \Bbb{N}$$ $$\implies (e^{n})^{-n} \le e^{-n}$$

$$\implies\sum_{n=1}^{\infty}\frac{1}{{e^{n^2}}} \le \sum_{n=1}^{\infty}\frac{1}{{e^{n}}} $$ $$\sum_{n=1}^{\infty}\frac{1}{{e^{n}}} = \sum_{n=1}^{\infty}(\frac{1}{e})^{n}=\frac{\frac{1}{e}}{1-\frac{1}{e}}=\frac{1}{e-1}$$ Therefore the series in question converges. I also thought about showing this in another way using the integral test. I have only read about the error function briefly so it is likely I did in fact make a mistake here: $$\frac{2}{\sqrt{\pi}}\sum_{n=2}^{\infty}\frac{1}{{e^{n^2}}}\le \frac{2}{\sqrt{\pi}}\int_1^\infty{e^{-t^{2}}}dt =\operatorname{erfc}(1) = 1-\operatorname{erf}(1)$$ $$\implies \sum_{n=2}^{\infty}\frac{1}{{e^{n^2}}} \le \frac{\sqrt{\pi}}{2}(1-\operatorname{erf}(1))$$ Adding the first term back into the series we see: $$\sum_{n=1}^{\infty}\frac{1}{{e^{n^2}}} = \frac{1}{e} + \sum_{n=2}^{\infty}\frac{1}{{e^{n^2}}} \le \frac{1}{e} + \frac{\sqrt{\pi}}{2}(1-\operatorname{erf}(1))$$ And since $\operatorname{erf}(1)$ and $\frac{1}{e}$ are both finite-valued this shows the series converges by the integral test.

My professor told me both of these are wrong, but I am struggling to see why. Specifically that the statement $e^{n^{2}}=(e^{n})^{n}\ge e^{n} \ ,\ \forall \ n \in \Bbb{N}$ is incorrect. She also said that $e^{-x^{2}}$ is non-integrable, so using the error function is not valid. I am confused why I can't use the error function if it's ok to use other transcendental functions like log(x), exp(x), etc.

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    $\begingroup$ I'm on your side on this one, both approaches seem correct. How could $x \mapsto e^{-x^2}$ not be integrable? Maybe she meant to say that you didn't give her the solution she wanted? $\endgroup$
    – Bruno B
    Commented Mar 19 at 18:15
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    $\begingroup$ Looks correct. But, is the error function something that's come up in class or in the text? If not, it's not valid to use unless you yourself define it and prove any properties about it you are using. $\endgroup$ Commented Mar 19 at 18:25
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    $\begingroup$ The only thing I can think of for the first one is that the prof didn't approve of the way you said it. Strictly speaking, it's enough to note that $n^2\geq n$ for $n\geq 0$ and thus $e^{n^2}\geq e^{n}$ for all $n$. (But this seems pretty flimsy.) $\endgroup$ Commented Mar 19 at 18:55
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    $\begingroup$ I can see some phrasing problems....for instance, you don't really argue that the geometric series converges, you just plug the ratio into a formula. But that doesn't work. I mean, if you had $\sum 2^n$ you could still plug $2$ into the formula, getting $\frac 1{1-2}=-1$, so would you conclude that $\sum 2^n$ converges? But, on balance, I'd certainly say your first approach was fine. $\endgroup$
    – lulu
    Commented Mar 19 at 19:53

2 Answers 2

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First, I agree with your ideas and what you have done. Both methods make sense. I don't quite understand why your prof didn't agree with them.

Second, there is a third way to do it.

$\textbf{Theorem:}$ Whenever $p > 1$, the series $\sum_{n = 1}^{\infty} \frac{1}{n^p}$ converges.

One famous example for this, is

$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$

Given this example, which must appear in almost every Calculus textbook, and given that

$$e^{n^2} > n^2$$

which implies

$$\frac{1}{e^{n^2}} < \frac{1}{n^2}$$

we have

$$0 < \sum_{n=1}^{\infty} \frac{1}{e^{n^2}} < \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$

By comparison property, we have $\sum_{n=1}^{\infty} \frac{1}{e^{n^2}}$ converges.

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    $\begingroup$ I like this, it’s much more concise. Thanks for pointing it out! $\endgroup$
    – bwootton
    Commented Mar 19 at 21:27
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We have $$\sqrt[n]{e^{-n^2}}=e^{-n}\to 0$$ By the Cauchy root test the series is convergent.

I am pretty sure that your professor had this solution in mind

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