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How do I evaluate this sum: $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2n!}$$

Note: The series converges by the ratio test. I have tried to use this sum:$$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}= \ln (2) $$ but I didn't succeed. Might there be others techniques which I don't know?

Thank you for any help

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  • $\begingroup$ Have you tried to differentiate you series for $\ln(x)$? $\endgroup$
    – Surb
    Commented Dec 14, 2015 at 23:03
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    $\begingroup$ A better idea would be to start with the series for $e^x-1$ instead of the series for $\ln x$, as that would give you the $n!$ in the numerator. Of course, it may be futile since Wolfram Alpha doesn't give a nice closed form. $\endgroup$
    – JimmyK4542
    Commented Dec 14, 2015 at 23:07
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    $\begingroup$ I can turn the sum into an integral like this: $$\int_0^1 du \, \int_0^1 dv \, \frac{1-e^{-u v}}{u v} $$ Not sure what good this does. $\endgroup$
    – Ron Gordon
    Commented Dec 14, 2015 at 23:22
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    $\begingroup$ Setting $f(x)=\displaystyle\sum_{n\ge1}\frac{x^{n+1}}{n^2n!}$, you can differentiate twice and come up with the second order ODE$$x^2f''(x)-xf'(x)+f(x)=x(e^x-1)$$which you can then try approaching with a power series solution via the Frobenius method. Mathematica returns a solution that agrees with the exact value of the series in terms of the generalized hypergeometric function. $\endgroup$
    – user170231
    Commented Dec 14, 2015 at 23:25
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    $\begingroup$ There is no reason to suppose it has a nice closed form, but it converges faster than an exponential function and the sum of the first $15$ terms show it is marginally greater than $0.8912127981113$ $\endgroup$
    – Henry
    Commented Dec 15, 2015 at 0:14

1 Answer 1

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We have that $$ \frac{1}{n^2}=\int_{0}^{1}(-\log x)\,x^{n-1}\,dx \tag{1}$$ hence: $$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2 n!}=\int_{0}^{1}(-\log x)\sum_{n\geq 1}\frac{(-1)^{n-1} x^{n-1}}{n!}\,dx =\int_{0}^{1}\frac{1-e^{-x}}{x}(-\log x)\,dx\tag{2}$$ and $$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2 n!}=\frac{d}{d\alpha}\left.\int_{0}^{1}\frac{1-e^{-x}}{x^\alpha}\,dx\,\right|_{\alpha=1}\tag{3}$$ is a value of the derivative of a sum of an exponential integral function and a $\Gamma$ function.
The series definition directly implies that $$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2 n!}= \phantom{}_3 F_3\left(1,1,1;2,2,2;-1\right).\tag{4}$$

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