Proving $3<\pi<4$

Partial fraction decomposition of the given sum is $\large \sum_{n=1}^{\infty} \frac{1}{n^3(n+1)^3}=\sum_{n=1}^{\infty} \left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)-3\sum_{n=1}^{\infty} \left(\frac{1}{n^2}+\frac{1}{(n+1)^2}\right)+6\sum_{n=1}^{\infty} \left(\frac{1}{n}-\frac{1}{n+1}\right)$

The first and the third partial sums telescope to $1$ and the second sum equals $2\zeta(2)-1=\dfrac{\pi^2}{3}-1$ . Hence the sum is $1-3\left(\dfrac{\pi^2}{3}-1\right)+6=10-\pi^2$ . Therefore $a=10,b=2$ making the answer $\boxed{12}$ .

Before we start to find the closed form for infinite series we shall start with Bonus problem.

I happen to derive this result while proving Charming inequality/approximation in maths stack exchange.

Preliminaries:

Consider the set $S=\left\{X_n= \frac{1}{n^3(n+1)^3} : n\in\mathbb {N} \right\}$ and here we show that set $S$ is bounded set with lower and bounds $0$ and $\frac{1}{8}$ respectively. Note that for $n\geq 1$ the $\begin{aligned} S_{n+1}-S_n & =\frac{1}{(n+1)^3}\left[\frac{1}{(n+2)^3}-\frac{1}{n^3}\right]\\ & =\frac{1}{(n+1)^3}\left[\frac{n^3-(n+2)^3}{n^3(n+2)^2}\right]\cdots(1)\end{aligned}$ since for all $n>1$ , $n< n+2\implies n^3-(n+2)^3<0$ from $(1)$ it follows that $S_{n+1}-S_n <0$ implies the sequence $X_n$ contained in set $S$ is a decreasing sequence and thus $\begin{aligned} \operatorname{sup}\left\{X_n : n\in \mathbb{N}\right\}&=\frac{1}{8}<1\\ \operatorname {inf}\left\{X_n:n\in\mathbb{N} \right\}&=0\end{aligned}$ Therefore, $0 < X_n \leq \frac{1}{8} <1$ . Futher, $n^3(n+1)^3> n(n+1)=Y_n$ and hence $0 <\sum_{n\geq 1} X_n < \sum_{n\geq 1} (Y_n)^{-1}=1\cdots(2)$ since we have telescoping series as $\displaystyle \sum_{n\geq 1} (Y_n)^{-1} =\sum_{n\geq 1} \left(\frac{1}{n}-\frac{1}{n+1}\right)=1$

Proof

$\begin{aligned} \sum_{n\geq 1} X_n & =\sum_{n\geq 1} \left(\frac{1}{Y_n}\right)^3\\&=\sum_{n \geq 1}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)\\&-\sum_{n\geq 1}\frac{3}{(Y_n)^2}=\zeta(3)-\zeta(3)+1\\& 1-3\sum_{n\geq 1} \left(\frac{1}{n^2}+\frac{1}{(n+1)^2}\right)\\&-2\sum_{n\geq 1}\frac{1}{(Y_n)}= 1-6\zeta(2)+9 \\&= 10-\pi^2\end{aligned}$ and thus from $(2)$ $\sum_{n\geq 1}X_n >0\implies \pi <\sqrt{10}$ since we want the closest bounding for $\pi$ so we find the closest perfect square number of $10$ which is as $9-\pi^2 < 10-\pi^2<16-\pi^2$ which implies $3< \pi < 4$ .