Sum of Every Power's Reciprocal

The sum is equal to $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2} + \frac{1}{n^3}+\frac{1}{n^4}...=\sum_{n=2}^{\infty}\frac{1}{n^2}.\frac{n}{n-1}=\sum_{n=2}^{\infty} \frac{1}{n.(n-1)}$ . But $\frac{1}{(n-1).n}=\frac{1}{n-1} -\frac{1}{n}$ and $\displaystyle\lim_{n \rightarrow \infty } \frac{1}{n}=0$ .So the sum is equal to $\frac{1}{2-1}=\boxed1$