# On a series involving harmonic numbers

Finding $$\sum^{\infty}_{n=1}\bigg[\bigg(\frac{(-1)^{n+1}}{n+1}\bigg)\bigg(\sum^{n}_{r=1}\frac{1}{r}\bigg)\bigg]$$

Try: Let $\displaystyle H_{n} =\sum^{n}_{r=1}\frac{1}{r},$ then series $\displaystyle \sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n+1}\cdot H_{n}=\frac{H_{1}}{2}-\frac{H_{2}}{3}+\frac{H_{3}}{4}\cdots\cdots$

could some help me to solve it, thanks

My solution (sorry, didn't have time to write it up earlier) isn't that short, but rather straightforward: I use the integral representations $$H_n=\int^1_0\frac{1-t^n}{1-t}\,dt$$ and $$\frac1{n+1}=\int^1_0u^n\,du,$$ so we have \begin{align}\sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n+1}\,H_{n}&=\int^1_0\int^1_0\sum^\infty_{n=1}(-1)^{n+1}\frac{u^n-u^nt^n}{1-t}\,dt\,du\\&=\int^1_0\int^1_0\left(\frac{u}{1+u}-\frac{ut}{1+ut}\right)\frac1{1-t}\,dt\,du\\&=\int^1_0\int^1_0\frac{u}{(1+u)(1+ut)}\,dt\,du\\&=\int^1_0\frac{\ln(1+u)}{1+u}\,du=\frac12\,\ln^22\end{align}
Summation by parts is a good way to go, generating functions another one. We have $$-\log(1-x) = \sum_{n\geq 1}\frac{x^n}{n}\tag{1}$$ $$\frac{-\log(1-x)}{1-x} = \sum_{n\geq 1} H_n\,x^n\tag{2}$$ $$\sum_{n\geq 1}\frac{H_n}{n+1}\,z^{n+1} = \int_{0}^{z}\frac{-\log(1-x)}{1-x}\,dx = \frac{1}{2}\log^2(1-z)\tag{3}$$ and by applying $\lim_{z\to -1^+}$ to both sides of $(3)$ you may easily recover Professor Vector's claim in the comments.