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I need to show, using the comparison test, that $\sum\limits_{n = 1}^{+\infty} e^{-\sqrt{n + 1}}$ converges, but I can't come up with a larger convergent series.

Thanks.

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You may try $\sum{\frac{1}{n^2}}$, for example, if you know that $\sqrt{n+1}$ is larger than $2\log{n}$ for sufficiently large $n$. –  user20266 Jun 4 '12 at 15:07
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up vote 5 down vote accepted

Since $\mathrm e^u=\sum\limits_{k=0}^{+\infty}\frac{u^k}{k!}\gt\frac{u^4}{4!}$ for every $u\gt0$, one has $\mathrm e^{-\sqrt{n+1}}\lt\frac{4!}{(n+1)^2}$ hence the series converges.

A similar argument shows that the series $\sum\limits_{n}\mathrm e^{-n^a}$ converges for every $a\gt0$.

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Would you mind explaining your answer in more detail? I understand $e^u = \sum\limits_{n = 0}^{+\infty} \frac{u^n}{n!}$, but $\frac{u^4}{4!}$ and $\frac{4!}{{(n + 1)}^2}$ come from? –  David Robert Jones Jun 4 '12 at 15:42
    
In the series you understand, all the terms are positive hence one term (here $\frac{u^4}{4!}$) is less than the sum $\mathrm e^u$ of the series. Write this as $\frac{4!}{u^4}\gt\mathrm e^{-u}$ and use it for $u=\sqrt{n+1}$. –  Did Jun 4 '12 at 15:46
    
got it! very impressive. –  David Robert Jones Jun 4 '12 at 16:51
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