$\begin{equation} \text { If } s_n=\sum_{r=0}^n T_r=\frac{(2 n-1)(2 n+1)(2 n+3)(2 n+5)}{64} \text { then find } \operatorname{Lim}_{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{T_r}= \end{equation}$ (22 Jan, Shift I, Memory Based)

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YESPYQ · Accepted Answer

Correct answer: a) $\frac{2}{3}$. $\begin{aligned} & T_n=S_{n-} S_{n-1} \ & =\frac{(2 n-1)(2 n+1)(2 n+3)(2 n+5)-(2 n-3)(2 n-1)(2 n+1)(2 n+3)}{64} \ & T_n=\frac{(2 n-1)(2 n+1)(2 n+3)}{8} \ & \frac{1}{T_n}=\frac{8}{(2 n-1)(2 n+1)(2 n+3)} \ & \frac{1}{T_n}=2\left(\frac{1}{(2 n-1)(2 n+1)}-\frac{1}{(2 n-1)(2 n+3)}\right) \ & \sum_{r=1}^n \frac{1}{T_r}=2\left(\frac{1}{1 \times 3}-\frac{1}{(2 n-1)(2 n+3)}\right) \ & \operatorname{Lim}_{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{T_r}=\frac{2}{3} \end{aligned}$

\(\begin{equation} \text { If } s_n=\sum_{r=0}^n T_r=\frac{(2 n-1)(2 n+1)(2 n+3)(2 n+5)}{64} \text { then find } \operat…

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