Show that the following series is less than $4 \pi^2 / 3$. Show: For any $k = 0,1,2,...$,
$$ \sum_{i=0}^{i=k} \frac{(k+1)^2}{(i+1)^2 (k-i+1)^2} \leq \frac{4 \pi^2}{3}. $$

Hint: 
1) From $(\frac{a+b}{2})^2\le\frac{a^2+b^2}{2}$ verify that $(k+1)^2 < 2[(i+1)^2+(k-i+1)^2]$,
2) $\sum_{n\ge 1}\frac{1}{n^2} = \frac{\pi^2}{6}$.
