Compare Infinite Sums

$A=\sum\limits_{n=2}^{\infty}(\sum\limits_{m=2}^{\infty}\left(\frac{1}{n}\right)^m)=\sum\limits_{n=2}^{\infty}\frac{1}{n^2}\cdot\frac{1}{1-\frac{1}{n}}=\sum\limits_{n=2}^{\infty}\frac{1}{n^2}\cdot\frac{n}{n-1}=\sum\limits_{n=2}^{\infty}\frac{1}{n\left(n-1\right)}=B$

where I used $\sum\limits_{n=2}^{\infty}a^n=a^2\sum\limits_{n=0}^{\infty}a^n=a^2\cdot\frac{1}{1-a}$ for $\left|a\right|<1$

A.

\begin{aligned} \sum_{\substack{n>1\\ m>1}}^{\infty} \frac{1}{n^{m}} &=(\frac{1}{2^{2}} + \frac{1}{2^{3}} + \frac{1}{2^{4}} + \cdots) + (\frac{1}{3^{2}} + \frac{1}{3^{3}} + \frac{1}{3^{4}} + \cdots) + (\frac{1}{4^{2}} + \frac{1}{4^{3}} + \frac{1}{4^{4}} + \cdots) +\cdots \\ &=(\frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \frac{1}{2^{4}} + \cdots) - \frac{1}{2} + (\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \frac{1}{3^{4}} + \cdots) - \frac{1}{3} + (\frac{1}{4} + \frac{1}{4^{2}} + \frac{1}{4^{3}} + \frac{1}{4^{4}} + \cdots) - \frac{1}{4}+\cdots \\\\ &=1-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \cdots \\\\\\ \end{aligned}

B.

$\begin{aligned} {\displaystyle \sum_{n=1}^{\infty} \frac{1}{n(n+1)}} &=\frac{1}{1\times2} + \frac{1}{2\times3} + \frac{1}{3\times4} + \cdots \\ &=1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \cdots \\\\ \end{aligned}$

Therefore, A and B are equal .