Sum of the Given Series (4)

$\begin{aligned} S & = \sum_{n=1}^\infty \frac 1{n^2(n+1)^2} & \small \color{#3D99F6} \text{By partial fraction decomposition} \\ & = \sum_{n=1}^\infty \left(- \frac 2n + \frac 2{n+1} + \frac 1{n^2} + \frac 1{(n+1)^2} \right) & \small \color{#3D99F6} \text{(see note)} \\ & = - 2 + \sum_{n=1}^\infty \frac 1{n^2} + \sum_{n=1}^\infty \frac 1{(n+1)^2} \\ & = -2 + \sum_{n=1}^\infty \frac 1{n^2} + \sum_{n=1}^\infty \frac 1{n^2} - 1 \\ & = 2 \sum_{n=1}^\infty \frac 1{n^2} - 3 \\ & = 2 {\color{#3D99F6}\zeta(2)} - 3 & \small \color{#3D99F6} \text{where }\zeta (n) = \sum_{k=1}^\infty \frac 1{k^n} \text{ is Riemann} \\ & = \frac {\pi^2}3 - 3 & \small \color{#3D99F6} \text{zeta function and }\zeta (2) = \frac {\pi^2}6. \end{aligned}$

Therefore, $A+B+C=2+3+3=\boxed{8}$ .

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Note: Let $\frac 1{n^2(n+1)^2} = \frac An + \frac B{n+1} + \frac C{n^2} + \frac D{(n+1)^2}$ . Multiplying both sides by $n^2(n+1)^2$ :

$\small \begin{aligned} An(n+1)^2 + Bn^2(n+1) + C(n+1)^2 + Dn^2 & = 1 & \color{#3D99F6} \text{Putting }n=0 \\ \implies C & = 1 & \color{#3D99F6} \text{Putting }n=-1 \\ \implies D & = 1 & \color{#3D99F6} \text{Putting }n=1 \\ \implies 2A+B & = -2 & \color{#3D99F6} \text{Putting }n=2 \\ \implies 3A+2B & = -2 \\ \implies A = -2 \implies B & = 2 \end{aligned}$

$S= \sum_{n=1}^{\infty} \dfrac{1}{n^2(n+1)^2} =\sum_{n=1}^{\infty}\left(\dfrac{1}{n(n+1)}\right)^2 \qquad {\color{#3D99F6}\text{see OR}}\\ S =\sum_{n=1}^{\infty} \left(\underbrace{\dfrac{1}{n^2} -\overbrace{2\left( \dfrac{1}{n}-\dfrac{1}{n+1} \right)}^{\text{telescoping sum}}+\dfrac{1}{(n+1)^2}}_{\text{partial fraction decomposition}}\right) \\S = \dfrac{\pi^2}{6}- 2 +\dfrac{\pi^2}{6} - 1 =\boxed{ \dfrac{\pi^2}{3} -3} \$

Hence, $A+B+C = 2+3+3 =8$

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OR $S=\sum_{n=1}^{\infty}\left(\dfrac{1}{n(n+1)}\right)^2 = \sum_{n=1}^{\infty} \left(\dfrac{1}{n} -\dfrac{1}{n+1}\right)^2 = \dfrac{1}{n^2} -\dfrac{2}{n(n+1)} +\dfrac{1}{(n+1)^2}$