Special summation 2

$\quad \text{The sum} \\= \displaystyle\sum_{n=1}^\infty \dfrac{1}{n^3(n+1)^3} \\= \displaystyle\sum_{n=1}^\infty \left[\dfrac{1}{n(n+1)}\right]^3 \\= \displaystyle\sum_{n=1}^\infty \left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)^3 \\= \displaystyle\sum_{n=1}^\infty \left[\dfrac{1}{n^3}-\dfrac{3}{n^2(n+1)}+\dfrac{3}{n(n+1)^2}-\dfrac{1}{(n+1)^3}\right] \\= \displaystyle\sum_{n=1}^\infty \left[\dfrac{1}{n^3}-\dfrac{1}{(n+1)^3}\right] -3\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2(n+1)^2} \\= 1-3\left(\dfrac{\pi^2}{3}-3\right) \\= 10-\pi^2 \\ \therefore A+B+C=-1+2+10=\boxed{11.000}$

*In the fifth step, I use the answer in Special summation 1 .

$\begin{aligned} S & = \sum_{n=1}^\infty \frac 1{n^3(n+1)^3} & \small \blue{\text{By partial fraction decomposition}} \\ & = \sum_{n=1}^\infty \left(\frac {6n^2-3n+1}{n^3} - \frac {6n^2+15n+10}{(n+1)^3}\right) \\ & = \sum_{n=1}^\infty \left(\frac 6n - \frac 3{n^2} + \frac 1{n^3} - \frac 6{n+1} - \frac 3{(n+1)^2} - \frac 1{(n+1)^3} \right) \\ & = 6 \sum_{n=1}^\infty \left(\frac 1n - \frac 1{n+1}\right) - 6 \blue{\sum_{n=1}^\infty \frac 1{n^2}} + 3\left(\frac 1{1^2}\right) + \sum_{n=1}^\infty \left(\frac 1{n^3} - \frac 1{(n+1)^3}\right) & \small \blue{\text{Riemann zeta function }\zeta (s) = \sum_{k=1}^\infty \frac 1{k^s}} \\ & = 6 - 6 \blue{\zeta (2)} + 3 + 1 = 10 - \pi^2 & \small \blue{\text{and }\zeta (2) = \frac {\pi^2}6} \end{aligned}$

Therefore, $A+B+C = - 1 + 2 + 10 = \boxed{11.000}$ .

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Reference:

• Partial fraction decomposition
• Riemann zeta function