Let $X$ have pmf
\[
p(x)=\begin{cases}
\frac{k}{x^2}, & x=1,2,3,\ldots\\
0, & \text{otherwise}
\end{cases}
\]
where $k$ is chosen so that $\sum_{x=1}^{\infty}\frac{k}{x^2}=1$. So the expected value of $X$ is
\[
\mu=E(X)=\sum_{x=1}^{\infty}x\cdot\frac{k}{x^2}=k\sum_{x=1}^{\infty}\frac{1}{x}
\]
What is the value of $k$?

Answer: The value of $k$ is equal to $\frac{6}{\pi^2}$.

Please see the Question 20, or search $\frac{\pi^2}{6}$ in this site.