The series $\sum\limits_{n=1}^{\infty}\frac{n!}{n^n}$ is clearly positive and decreasing, but how does one go about integrating $\int\frac{x!}{x^x} dx$ ?
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How would you define x! ? Do you mean $\Gamma(x)$? – Vim Sep 13 '15 at 11:02
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x! = x(x-1)(x-2)...(3)(2)(1) – bard Sep 13 '15 at 11:05
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2it is invalid when x is not an integer. – Vim Sep 13 '15 at 11:06
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But how would you integrate that ?. I mean x! – Sep 13 '15 at 11:07
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'The series...is decreasing' That's not true: the series is increasing because every term is positive. What you mean is the sequence $\frac{k!}{k^k}$ is decreasing – Alex Sep 13 '15 at 12:49
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1I'd use the Ratio Test for this series instead. – user84413 Sep 13 '15 at 21:58
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Linking this one. – metamorphy Mar 29 '23 at 03:59
2 Answers
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$\lim\limits_{n \to \infty}\sqrt[n]{\dfrac{n!}{n^n}}\sim \lim\limits_{n \to \infty} \dfrac{\dfrac{n}{e}}{n}=\dfrac{1}{e}<1$
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A very rough way of showing convergence is use Stirling's expansion of the numerator. Since $k! \sim (\frac{k}{e})^k \sqrt{2 \pi k}$ the summand becomes $a_k \sim \frac{\sqrt{2 \pi k}}{e^k}$ which you can then compare to the integral, $\Gamma(\frac{3}{2})$.
This, I repeat, is a pretty rough and not very elegant way of solving the problem. There are better ones too.
Alex
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