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Test the convergence of the series

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What tests have you tried? What can you say about the numerator? – Calvin Lin Jan 29 '13 at 7:53
I have tried cauchy root test, it gives no conclusion. – asimath Jan 29 '13 at 7:55
You will find the Stirling approximation to the factorial works very nicely. One could get away with somewhat less. – André Nicolas Jan 29 '13 at 7:58
up vote 12 down vote accepted

Observe that by Stirling's approximation we always have $$n!\le e\sqrt{n}\left(\frac{n}{e}\right)^n$$ and so we see that $$\sum_{n=1}^\infty\frac{\left(1+\frac{n}{e}\right)^n}{n!}\ge \sum_{n=1}^\infty\frac{\left(\frac{n}{e}\right)^n}{e\sqrt{n}\left(\frac{n}{e}\right)^n}=\frac{1}{e}\sum_{n=1}^\infty\frac{1}{\sqrt{n}}=\infty$$ thus the series diverges.

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we have $$\left(\frac{n}{\mathrm{e}}\right)\leq \left(1+\frac{n}{\mathrm{e}}\right)$$
And so we have$$\frac{{\left(\frac{n}{\mathrm{e}}\right)}^n}{n\mathrm !}\leq \frac{{\left(1+\frac{n}{\mathrm{e}}\right)}^n}{n\mathrm !}$$
Now the series $\sum_{n=1}^{\infty}\frac{{\left(\frac{n}{\mathrm{e}}\right)}^n}{n\mathrm !}$ is divergent by logarithmic test. So the given series is divergent.

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I think doing in this way would be much easier. – asimath Jan 29 '13 at 8:20
Why are you asking this question if you already have a solution? -2. – Benjamin Dickman Jan 29 '13 at 8:22
I have found it by seeing the previous answer. To be more precise by seeing 2nd step of it. – asimath Jan 29 '13 at 8:23

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