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I am trying to find an upper bound to this expression

$$ \sum_{n=1}^\infty \frac{2^n - n^3}{\sqrt{n!}} $$

It is sure that the limit of the inner expression as it goes to infinity is 0 (which is neccesary, not enough), but I can't find a way to show that the sum exists.

klaufir
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  • Do you know Stirling's approximation for $n!$? (Interrobang not intended.) – Micah Jun 10 '13 at 06:15
  • Are you trying to compute an upper bound or simply show the sum converges? How good does the upper bound need to be, if that's what you're after? – Zach L. Jun 10 '13 at 06:16
  • I am trying to give some upper bound (maybe lower too if it converges from negative). The bounds don't have to very precise, just something with good reasoning behind it. – klaufir Jun 10 '13 at 06:18

1 Answers1

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For large $\,n\,$ we have Stirling's Approximation

$$n!\cong n^ne^{-n}\sqrt{2\pi n}$$

and from here

$$\frac{2^n-n^3}{\sqrt{n!}}\cong\frac{2^n-n^3}{n^{n/2}e^{-n/2}\sqrt[4]{2\pi n}}\le\left(\frac{2e}n\right)^n\frac{n^{n/2}}{e^{n/2}\sqrt[4]{2\pi n}}$$

and if you now apply Cauchy's $\,n$-th root test to the right side term, we get

$$\frac{2e}n\frac{\sqrt n}{\sqrt e(2\pi n)^{1/4n}}\xrightarrow[n\to\infty]{}0<1\;\ldots\ldots$$

DonAntonio
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