Alternating Euler Sum

Evaluate the sum:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\left ( 1-\frac{1}{2}+\cdots +\frac{(-1)^{n-1}}{n} \right )$$

Solution:


Let as denote as $\displaystyle \mathcal{H}^{-}_n=1-\frac{1}{2}+\cdots +\frac{(-1)^{n-1}}{n}$. Then the sum in question is written as:

$$\sum_{n=1}^{\infty}\frac{\left ( -1 \right )^{n-1}\mathcal{H}_n^{-}}{n}$$

Using the gen. function $\displaystyle \sum_{n=1}^{\infty}\mathcal{H}_n^{-1}x^n =\frac{\ln (1+x)}{1-x}$ it is easy to note that:

$$\frac{\ln (1+x)}{1-x}=\frac{\mathrm{d} }{\mathrm{d} x}{\rm Li}_2 \left ( \frac{1-x}{2} \right )+\frac{\ln 2}{1-x}$$

This means :

$$\sum_{n=1}^{\infty}\mathcal{H}_n^{-}x^n =\frac{\mathrm{d} }{\mathrm{d} x}{\rm Li}_2 \left ( \frac{1-x}{2} \right )+\frac{\ln 2}{1-x}$$

Integrating the above equation we get:

$$ \sum_{n=1}^{\infty}\frac{x^{n+1}\mathcal{H}_n^{-}}{n+1}={\rm Li}_2 \left ( \frac{1-x}{2} \right )-{\rm Li}_2 \left ( \frac{1}{2} \right )-\ln 2 \ln (1-x)$$

However, $\displaystyle \sum_{n=1}^{\infty}\frac{x^{n+1}\mathcal{H}_n^{-}}{n+1}=\sum_{n=1}^{\infty}\frac{x^n \mathcal{H}_n^{-}}{n}+{\rm Li}_2(-x)$ hence:

$$\sum_{n=1}^{\infty}\frac{x^n {\mathcal{H}_n^{-}}}{n}={\rm Li}_2 \left ( \frac{1-x}{2} \right )-{\rm Li}_2(-x)-{\rm Li}_2 \left ( \frac{1}{2} \right )-\ln 2 \ln (1-x)$$

Plugging in the previous equation $x=-1$ and multiplying everything with $-1$ we get that the series equals $\displaystyle \frac{\pi^2}{12}-\frac{\ln^2 2}{2}$.

and the exercise comes to an end.