Mathematical Reflections 2017, Issue 1 - Problem U397

Problem:
Let $T_n$ be the $n$-th triangular number. Evaluate $$\sum_{n \geq 1} \dfrac{1}{(8T_n-3)(8T_{n+1}-3)}$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Let $t_n=\dfrac{1}{(8T_n-3)(8T_{n+1}-3)}$.
Observe that $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \dfrac{1}{t_n}=(8T_n-3)(8T_{n+1}-3)&=&\left(8\dfrac{n(n+1)}{2}-3\right)\left(8\dfrac{(n+1)(n+2)}{2}-3\right)\\&=&(4n^2+4n-3)(4n^2+12n+5)\\&=&(2n-1)(2n+3)(2n+1)(2n+5). \end{array}$$
We get $$\renewcommand{\arraystretch}{2} \begin{array}{lll} t_n&=&\dfrac{1}{(2n-1)(2n+3)(2n+1)(2n+5)}\\&=&\dfrac{1}{8}\dfrac{1}{(2n-1)(2n+5)}-\dfrac{1}{8}\dfrac{1}{(2n+1)(2n+3)}\\&=&\dfrac{1}{48}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+5}\right)-\dfrac{1}{16}\left(\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\right)\\&=&\dfrac{1}{48}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)+\dfrac{1}{48}\left(\dfrac{1}{2n+3}-\dfrac{1}{2n+5}\right)-\dfrac{1}{24}\left(\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\right). \end{array}$$
So, $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \displaystyle \sum_{n \geq 1} t_n&=& \displaystyle \dfrac{1}{48}\sum_{n=1}^\infty \left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)+\dfrac{1}{48}\sum_{n=1}^\infty \left(\dfrac{1}{2n+3}-\dfrac{1}{2n+5}\right)-\dfrac{1}{24}\sum_{n=1}^\infty \left(\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\right) \\ &=&\dfrac{1}{48}+\dfrac{1}{48}\cdot\dfrac{1}{5}-\dfrac{1}{24}\cdot\dfrac{1}{3} \\ &=& \dfrac{1}{90}. \end{array}$$