300 Followers Problem - To my friends Swapnil and Sharky

First note that $S \lt -1 + \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^{2}},$ which converges by the p-test. Thus we will be able to rearrange terms of $S$ without affecting the sum.

Now the nth term in the given series is of the form $\dfrac{1}{n(n + 1)^{2}}.$

Decomposing this fraction, we let $\dfrac{1}{n(n + 1)^{2}} = \dfrac{A}{n} + \dfrac{B}{n + 1} + \dfrac{C}{(n + 1)^{2}}$

$\Longrightarrow 1 = A(n + 1)^{2} + Bn(n + 1) + Cn = (A + B)n^{2} + (2A + B + C)n + A.$

Equating like coefficients, we see that $A = 1,$ and that $A + B = 0 \Longrightarrow B = -A = -1.$ Then as $2A + B + C = 0,$ we find that $C = -(2A + B) = -(2 - 1) = -1.$

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

Now the first of these sums telescopes, leaving just the first element of the first term, namely $1.$ The second sum is the Basel problem solution . Thus

$S = 1 - \left(-1 + \dfrac{\pi^{2}}{6}\right) = 2 - \dfrac{\pi^{2}}{6} = 0.3550659332..... \Longrightarrow \lfloor 10000 \times S \rfloor = \boxed{3550}.$