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#### Alexmahone

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- Jan 26, 2012

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Do the following series converge or diverge?

a) $\displaystyle \sum\frac{1}{n!}\left(\frac{n}{e}\right)^n$

b) $\displaystyle\sum\frac{(-1)^n}{n!}\left(\frac{n}{e}\right)^n$

a) $\displaystyle\frac{1}{n!}\left(\frac{n}{e}\right)^n\approx\frac{1}{\sqrt{2\pi n}}$ (Stirling’s formula)

$\displaystyle \sum\frac{1}{n!}\left(\frac{n}{e}\right)^n\approx \sum\frac{1}{\sqrt{2\pi n}}$

Since $\displaystyle\sum\frac{1}{\sqrt{2\pi n}}$ diverges, $\displaystyle \sum\frac{1}{n!}\left(\frac{n}{e}\right)$ also diverges.

b) $\displaystyle\frac{(-1)^n}{n!}\left(\frac{n}{e}\right)^n\approx\frac{(-1)^n}{\sqrt{2\pi n}}$ (Stirling’s formula)

$\displaystyle \sum\frac{(-1)^n}{n!}\left(\frac{n}{e}\right)^n\approx \sum\frac{(-1)^n}{\sqrt{2\pi n}}$

By Leibniz's test for alternating series, $\displaystyle\sum\frac{(-1)^n}{\sqrt{2\pi n}}$ converges.

So, $\displaystyle\sum\frac{(-1)^n}{n!}\left(\frac{n}{e}\right)^n$ also converges.

a) $\displaystyle \sum\frac{1}{n!}\left(\frac{n}{e}\right)^n$

b) $\displaystyle\sum\frac{(-1)^n}{n!}\left(\frac{n}{e}\right)^n$

__My attempt:__a) $\displaystyle\frac{1}{n!}\left(\frac{n}{e}\right)^n\approx\frac{1}{\sqrt{2\pi n}}$ (Stirling’s formula)

$\displaystyle \sum\frac{1}{n!}\left(\frac{n}{e}\right)^n\approx \sum\frac{1}{\sqrt{2\pi n}}$

Since $\displaystyle\sum\frac{1}{\sqrt{2\pi n}}$ diverges, $\displaystyle \sum\frac{1}{n!}\left(\frac{n}{e}\right)$ also diverges.

b) $\displaystyle\frac{(-1)^n}{n!}\left(\frac{n}{e}\right)^n\approx\frac{(-1)^n}{\sqrt{2\pi n}}$ (Stirling’s formula)

$\displaystyle \sum\frac{(-1)^n}{n!}\left(\frac{n}{e}\right)^n\approx \sum\frac{(-1)^n}{\sqrt{2\pi n}}$

By Leibniz's test for alternating series, $\displaystyle\sum\frac{(-1)^n}{\sqrt{2\pi n}}$ converges.

So, $\displaystyle\sum\frac{(-1)^n}{n!}\left(\frac{n}{e}\right)^n$ also converges.

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