If $\sum _{\mathrm{r}=1}^{\mathrm{n}} {\mathrm{T}}_{\mathrm{r}}=\frac{(2\mathrm{n}-1)(2\mathrm{n}+1)(2\mathrm{n}+3)(2\mathrm{n}+5)}{64},$then $\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \sum _{\mathrm{r}=1}^{\mathrm{n}} \left(\frac{1}{{\mathrm{T}}_{\mathrm{r}}}\right)$ is equal to :

$\begin{array}{r}{T}_n=S_n-S_{n-1}\end{array}$

$\begin{array}{r}\Rightarrow T_n=\frac{1}{8}(2n-1)(2n+1)(2n+3)\end{array}$

$\begin{array}{r}\Rightarrow \frac{1}{T_n}=\frac{8}{(2n-1)(2n+1)(2n+3)}\end{array}$

$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim} \sum _{r=1}^n \frac{1}{T_r}=\underset{\mathrm{n}\to \mathrm{\infty }}{lim} 8\sum _{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{(2\mathrm{n}-1)(2\mathrm{n}+1)(2\mathrm{n}+3)}\end{array}$

$\begin{array}{r}=\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \frac{8}{4}\sum \left(\frac{1}{(2\mathrm{n}-1)(2\mathrm{n}+1)}-\frac{1}{(2\mathrm{n}+1)(2\mathrm{n}+3)}\right)\end{array}$

$\begin{array}{r}=\underset{\mathrm{n}\to \mathrm{\infty }}{lim} 2\left[\left(\frac{1}{1.3}-\frac{1}{3.5}\right)+\left(\frac{1}{3.5}-\frac{1}{5.7}\right)+\dots \right]\end{array}$

$=\frac{2}{3}$