The infinite sum $\sum_{n=1}^\infty (-1)^{n+1} \frac{2n-1}{n^2-n+1}$ My (rather old) version of Mathematica cannot compute 
$$\sum_{n=0}^\infty (-1)^{n+1}\frac{2n-1}{n^2-n+1}$$ 
other than re-writing it as a hypergeometric function as follows:
$$\mbox{HypergeometricPFQ}\left[\left\{\frac{1}{2}, 1, -(-1)^{1/3}, (-1)^{2/3}\right\}, \left\{-\frac{1}{2}, (-1)^{1/3}, -(-1)^{2/3}\right\}, -1\right]$$
But strangely enough, my old Mathematica has no trouble finding 
$$\sum_{n=1}^\infty (-1)^{n+1}\frac{2n-1}{n^2-n+1}=\frac{2\pi}{e^{\pi \sqrt{3}/2}+e^{-\pi \sqrt{3}/2}}$$ 
SAGE (with CoCalc) can't compute either version. I can't either. How does the old Mathematica do it?

Using partial fraction decomposition
$$\frac{2 n-1}{(n-a) (n-b)}=\frac{2 a-1}{(a-b) (n-a)}+\frac{1-2 b}{(a-b) (n-b)}$$ So
$$\sum_{n=1}^\infty (-1)^{n+1}\frac{2 a-1}{(a-b) (n-a)}=\frac{(2 a-1) \Phi (-1,1,1-a)}{a-b}$$
$$\sum_{n=1}^\infty (-1)^{n+1}\frac{1-2 b}{(a-b) (n-b)}=-\frac{(2 b-1) \Phi (-1,1,1-b)}{a-b}$$ where appears  the Hurwitz-Lerch transcendent function.
Now, using $a=(-1)^{1/3}=\frac{1+i \sqrt{3}}{2}$ and $b=-(-1)^{2/3}=\frac{1-i \sqrt{3}}{2}$ gives for the sum
$$S=\sum_{n=1}^\infty (-1)^{n+1} \frac{2n-1}{n^2-n+1}$$
$$S=\frac{1}{2} \left(-\psi \left(\frac{1}{4}-\frac{i \sqrt{3}}{4}\right)+\psi
   \left(\frac{3}{4}-\frac{i \sqrt{3}}{4}\right)-\psi
   \left(\frac{1}{4}+\frac{i \sqrt{3}}{4}\right)+\psi
   \left(\frac{3}{4}+\frac{i \sqrt{3}}{4}\right)\right)$$ which reduces to
$$S=\pi\,  \text{sech}\left(\frac{\sqrt{3} \pi }{2}\right)$$
