How do i evaluate this sum $\sum\limits_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2n!}$?

Solution 1:

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 $$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2 n!}= \phantom{}_3 F_3\left(1,1,1;2,2,2;-1\right).\tag{4}$$