Special summation 1

$\quad \text{The sum}\\=\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2(n+1)^2} \\ = \displaystyle\sum_{n=1}^\infty \left[\dfrac{1}{n(n+1)}\right]^2 \\ = \displaystyle\sum_{n=1}^\infty \left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)^2 \\= \displaystyle\sum_{n=1}^\infty \left[\dfrac{1}{n^2}-\dfrac{2}{n(n+1)}+\dfrac{1}{(n+1)^2}\right] \\ = 2\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2}-1-2\displaystyle\sum_{n=1}^\infty\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right) \\ =2\times \dfrac{\pi^2}{6}-1-2 \\=\dfrac{\pi^2}{3}-3 \\ \therefore A+B+C=\dfrac{1}{3}+2-3\approx\boxed{-0.667}$

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

Therefore, $A+B+C = \dfrac 13 + 2 - 3 = - \dfrac 23 \approx \boxed{-0.667}$ .

---

Reference:

• Partial fraction decomposition
• Riemann zeta function