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.
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.