# Convergenge or divergence of $\sum_{n=1}^\infty e^{-n^{2}}$ [closed]

I can't find the convergence or divergence of the following series by using aspect ratio test or comparison test. The series is:

$$\sum_{n=1}^\infty e^{-n^{2}}$$

Thanks.

## closed as off-topic by Martin R, Jam, metamorphy, mrtaurho, cmkJun 28 at 0:01

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• Given the range of answers below, can you show your work for your attempts? – Eric Towers Jun 25 at 1:17

Since $$e^{-n^2}\leq e^{-n}$$, the series converges by comparison with the geometric series $$\sum_{n=1}^\infty e^{-n}$$

Since$$(\forall n\in\mathbb N):\sqrt[n]{e^{-n^2}}=e^{-n}\to0,$$the series converges, by the root test.

Also by ratio test: $$\left|\frac{a_{n+1}}{a_n}\right| = \frac{1}{e^{2n+1}} \to 0 \text{ as n \to \infty}$$

Also by comparison test: $$e^{n^2} > n^2 \implies \frac{1}{e^{n^2}} < \frac{1}{n^2}$$

The series converges rather violently.

Since $$n^2 \ge n$$, $$e^{-n^2} \le e^{-n}$$, and the sum of that converges so your series also converges.

It converges also via the ratio test : $$\frac{e^{-(n+1)^2}}{e^{-n^2}} = e^{-2n - 1} \rightarrow 0$$, which is a limit of absolute value $$< 1$$.

Another way: Since $$e^{-n} > 0$$ for all $$n$$, $$\sum_{n=1}^N e^{-n^2} = \sum_{n=1}^N \prod_{k=1}^n e^{-n} < \sum_{n=1}^N \prod_{k=1}^n e^{-1} = \sum_{n=1}^N e^{-n} = \frac{1 - e^{-N}}{e - 1}.$$ Then as $$N \to \infty$$, $$\sum_{n=1}^\infty e^{-n^2} < \frac{1}{e-1}.$$