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Let $$a(n)=(n-1)!\frac{e^n}{n^{n-1/2}} - \sqrt{2\pi}$$ for n=1 to infinity. Does the sum of $a(n)$'s, i.e. $\sum_{n=1}^\infty a(n)$, converge?

DirkGently
  • 1,618

1 Answers1

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Stirling's formula is the hint.

$$n!\sim \sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n.$$

Your $a(n)+\sqrt{2\pi}=\dfrac{n!}{n}\left(\dfrac{e}{n}\right)^n\sqrt{n}\sim \sqrt{2\pi}.$

Finally, $a(n)\rightarrow_{n\to\infty}0.$

npisinp
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