If Σ_r=1^n T_r= (2 n-1)(2 n+1)(2 n+3)(2 n+5) 64, then lim _n → ∞ Σ_r=1^n ( 1 T_r ) is equal to :

$$\begin{aligned} & \mathrm{T}_{\mathrm{n}}=\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1} \\ & \Rightarrow \mathrm{~T}_{\mathrm{n}}=\frac{1}{8}(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3) \\ & \Rightarrow \frac{1}{\mathrm{~T}_{\mathrm{n}}}=\frac{8}{(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3)} \\ & \lim _{\mathrm{n} \rightarrow \infty} \sum_{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{\mathrm{~T}_{\mathrm{r}}}=\lim _{\mathrm{n} \rightarrow \infty} 8 \sum_{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3)} \\ & =\lim _{\mathrm{n} \rightarrow \infty} \frac{8}{4} \sum\left(\frac{1}{(2 \mathrm{n}-1)(2 \mathrm{n}+1)}-\frac{1}{(2 \mathrm{n}+1)(2 \mathrm{n}+3)}\right) \\ & =\lim _{\mathrm{n} \rightarrow \infty} 2\left[\left(\frac{1}{1.3}-\frac{1}{3.5}\right)+\left(\frac{1}{3.5}-\frac{1}{5.7}\right)+\ldots\right] \\ & =\frac{2}{3}\end{aligned}$$​Tn​=Sn​−Sn−1​⇒ Tn​=81​(2n−1)(2n+1)(2n+3)⇒ Tn​1​=(2n−1)(2n+1)(2n+3)8​n→∞lim​r=1∑n​ Tr​1​=n→∞lim​8r=1∑n​(2n−1)(2n+1)(2n+3)1​=n→∞lim​48​∑((2n−1)(2n+1)1​−(2n+1)(2n+3)1​)=n→∞lim​2[(1.31​−3.51​)+(3.51​−5.71​)+…]=32​​