How can we show that $1-\gamma=\sum_{n=1}^{\infty}{(-1)^{n-1}\over n(n+1)}\sum_{k=1}^{\infty}{1\over (k+1)k^n}$

Question

$$1-\gamma=\sum_{n=1}^{\infty}{(-1)^{n-1}\over n(n+1)}\sum_{k=1}^{\infty}{1\over (k+1)k^n}$$

How would we go about proving this series?

Answer 1

We have $$\sum_{n\geq1}\frac{\left(-1\right)^{n-1}}{n\left(n+1\right)}\sum_{k\geq1}\frac{1}{\left(k+1\right)k^{n}}=\sum_{k\geq1}\frac{1}{k+1}\sum_{n\geq1}\frac{\left(-1\right)^{n-1}}{n\left(n+1\right)k^{n}} $$ $$=\sum_{k\geq1}\frac{1}{k+1}\left(1-\left(k+1\right)\log\left(1+\frac{1}{k}\right)\right)=\sum_{k\geq1}\left(\frac{1}{k+1}-\log\left(1+\frac{1}{k}\right)\right)\tag{1} $$ $$=\sum_{k\geq1}\left(\frac{1}{k}-\log\left(1+\frac{1}{k}\right)+\frac{1}{k+1}-\frac{1}{k}\right)=\color{red}{\gamma-1}. $$ Note that we used the classic representation of the Euler-Mascheroni constant $$\sum_{k\geq1}\left(\frac{1}{k}-\log\left(1+\frac{1}{k}\right)\right)=\gamma $$ and in $(1)$ we used the Taylor series of $\log$ $$\log\left(1+x\right)=\sum_{n\geq1}\frac{\left(-1\right)^{n-1}}{n}x^{n}\Rightarrow\frac{x-\log\left(1+x\right)-x\log\left(1+x\right)}{x}=\sum_{n\geq1}\frac{\left(-1\right)^{n-1}}{n\left(n+1\right)}x^{n}. $$