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Suppose you are given a series with a factorial, is it appropriate to replace $n!$ with $\Gamma(n+1)$ and apply the integral test? It seems like even if there are other methods that work to show convergence better, this still might be potentially a good technique for approximating a series that doesn't fit into the category of trivial to evaluate.

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    How would you provide for (1) when you were dividing by $n!$ and (2) the fact that you would have to integrate it even after making the substitution. Do you know a simple anti-derivative for $\Gamma(x)$ or--if not--how to work around this in the integral test? – 2017-02-06
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    The basic integral test that $\sum_{k=1}^\infty f(k)$ and $\int_1^\infty f(x) \, dx$ converge/diverge together applies when $f$ is nonnegative and decreasing for $x \geqslant 1$. As long as this holds the appearance of the gamma function might not be a problem since it coincides with factorial at integers and is eventually monotone. You don't need to compute the integral -- only prove convergence. That may or may not be feasible depending on the situation. – 2017-02-06
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    I DO want to estimate the series though so I believe I'd need to evaluate the integral. – 2017-02-06
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    Also..... Can I apply L'hospital as well to the gamma function when dealing with the factorial? – 2017-02-06
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    I'm trying to evaluate $\sum_{n=1}^\infty \frac{n!}{n^n}$. I find the ratio test inconclusive ( unless I made an error) and using the root test (and wolfram) i have $\frac{(n!)^{1/n}}{n}\to e^{-1}$ but I don't know how to show this last fact and so I thought I could solve the original problem with the integral test and get a great estimate on the series too – 2017-02-06
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    $e^{-1} = \lim_{n \to \infty} (n!)^n /n = \lim_{x \to \infty} \Gamma(x+1))^{1/x} /x $ is true. Using LHR requires evaluating $\frac{d}{dx} \exp ( \log \Gamma (x+1) / x) = \Gamma(x+1)^{1/x} ( - \log \Gamma (x+1) /x^2 + \frac{d}{dx} \log \Gamma (x+1) /x)$ and finding the asymptotic behavior as $x \to \infty$. The root test shows the series converges since $e^{-1} < 1$. – 2017-02-06
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    The ratio test also works since $\frac{(n+1)!}{(n+1)^{n+1}} \frac{n^n}{n!} = 1/(1 + 1/n)^n \to e^{-1}$. – 2017-02-06
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    Embarrassingly easy to see now. Thank you – 2017-02-06

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