Mathematics - Sequence & Series Question with Solution | TestHub

$\begin{array}{rl}& {\mathrm{T}}_{\mathrm{n}}={\mathrm{S}}_{\mathrm{n}}-{\mathrm{S}}_{\mathrm{n}-1}\\ & \Rightarrow {\mathrm{T}}_{\mathrm{n}}=\frac{1}{8}(2\mathrm{n}-1)(2\mathrm{n}+1)(2\mathrm{n}+3)\\ & \Rightarrow \frac{1}{{\mathrm{T}}_{\mathrm{n}}}=\frac{8}{(2\mathrm{n}-1)(2\mathrm{n}+1)(2\mathrm{n}+3)}\\ & \underset{\mathrm{n}\to \infty }{\lim }\sum _{\mathrm{r}=1}^{\mathrm{n}}\frac{1}{{\mathrm{T}}_{\mathrm{r}}}=\underset{\mathrm{n}\to \infty }{\lim }8\sum _{\mathrm{r}=1}^{\mathrm{n}}\frac{1}{(2\mathrm{n}-1)(2\mathrm{n}+1)(2\mathrm{n}+3)}\\ & =\underset{\mathrm{n}\to \infty }{\lim }\frac{8}{4}\sum \left(\frac{1}{(2\mathrm{n}-1)(2\mathrm{n}+1)}-\frac{1}{(2\mathrm{n}+1)(2\mathrm{n}+3)}\right)\\ & =\underset{\mathrm{n}\to \infty }{\lim }2\left[\left(\frac{1}{1.3}-\frac{1}{3.5}\right)+\left(\frac{1}{3.5}-\frac{1}{5.7}\right)+\dots \right]\\ & =\frac{2}{3}\end{array}$