Ramanujan's Result

Prove that:

$$\sum_{k=1}^{\infty}\frac{1}{k^3 (k+1)^3}= 10 - \pi^2$$

Solution



We apply partial fractions. We note that:

$$\frac{1}{n^3 (n+1)^3}= \left ( \frac{1}{n^3} - \frac{1}{(n+1)^3} \right ) - \frac{3}{n^2} - \frac{3}{(n+1)^2} + 6 \left ( \frac{1}{n} - \frac{1}{n+1} \right )$$

Hence:

$$\begin{aligned}\sum_{n=1}^{\infty}\frac{1}{n^3 (n+1)^3}&= \sum_{n=1}^{\infty}\left[ \left ( \frac{1}{n^3} - \frac{1}{(n+1)^3} \right ) - \frac{3}{n^2} - \frac{3}{(n+1)^2} + 6 \left ( \frac{1}{n} - \frac{1}{n+1} \right ) \right]\\
&=  \sum_{n=1}^{\infty}\left ( \frac{1}{n^3} - \frac{1}{(n+1)^3} \right ) - 3 \sum_{n=1}^{\infty}\frac{1}{n^2} - 3\sum_{n=1}^{\infty}\frac{1}{(n+1)^2} \\
 & \quad \quad \quad \quad \qquad \qquad + 6\sum_{n=1}^{\infty}\left ( \frac{1}{n} - \frac{1}{n+1} \right ) \\
&= 7 - \frac{3 \pi^2}{6} - \frac{3\pi^2}{6} +3 \\
&=10 - \pi^2
\end{aligned}$$

which is the wanted result.