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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.

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

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  • 1
    $\begingroup$ Given the range of answers below, can you show your work for your attempts? $\endgroup$ – Eric Towers Jun 25 at 1:17
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Since $e^{-n^2}\leq e^{-n}$, the series converges by comparison with the geometric series $\sum_{n=1}^\infty e^{-n}$

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

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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: $$ e^{n^2} > n^2 \implies \frac{1}{e^{n^2}} < \frac{1}{n^2} $$

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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.

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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 $< 1$.

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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 $N \to \infty$, $$\sum_{n=1}^\infty e^{-n^2} < \frac{1}{e-1}.$$

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