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How would I go about using the limit comparison test on this series. I'm just not sure what value of 'bn' I would start with.. $$\sum_{n\geq1 }\frac{n!}{e^{n^{2}}}$$

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    $\begingroup$ I would rather use the Root Test. Ratio Test is good too. $\endgroup$ – André Nicolas Mar 12 '15 at 6:05
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If you want to use limit comparison, this is a way. By Stirling's approximation we have$$\lim_{n\rightarrow\infty}\frac{n!}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}}=1$$ so we have$$\lim_{n\rightarrow\infty}\frac{n!}{e^{n^{2}}}\frac{e^{n^{2}+n}}{\sqrt{2\pi n}n^{n}}=1$$ so we have to study$$\sqrt{2\pi}\sum_{n\geq1}\frac{n^{n+1/2}}{e^{n^{2}+n}}$$ now note$$e^{n^{2}+n}\geq n^{2n+1/2}$$ (you can take logs to see it), then$$\sqrt{2\pi}\sum_{n\geq1}\frac{n^{n+1/2}}{e^{n^{2}+n}}\leq\sqrt{2\pi}\sum_{n\geq1}\frac{1}{n^{n}}<\infty.$$

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