Show that the following series is less than $4 \pi^2 / 3$.

Question

Show: For any $k = 0,1,2,...$,

$$ \sum_{i=0}^{i=k} \frac{(k+1)^2}{(i+1)^2 (k-i+1)^2} \leq \frac{4 \pi^2}{3}. $$

Answer 1

Note that $$ \frac{1}{i+1}+\frac{1}{k-i+1}=\frac{k+2}{(i+1)(k-i+1)} $$ So, (because $(a+b)^2\leq 2a^2+2b^2$): $$ \frac{(k+2)^2}{(i+1)^2(k-i+1)^2}\leq\frac{2}{(i+1)^2}+\frac{2}{(k-i+1)^2} $$ Adding we get $$ \sum_{i=0}^{k}\frac{(k+2)^2}{(i+1)^2(k-i+1)^2} \leq 4\sum_{i=0}^{k}\frac{1}{(1+i)^2}\leq 4\zeta(2)=\frac{2\pi^2}{3} $$ So $$ \sum_{i=0}^{k}\frac{(k+1)^2}{(i+1)^2(k-i+1)^2} \leq \frac{(k+1)^2}{(k+2)^2}\frac{2\pi^2}{3}<\frac{2\pi^2}{3}< \frac{4\pi^2}{3} $$

Answer 2

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) $\sum_{n\ge 1}\frac{1}{n^2} = \frac{\pi^2}{6}$.