\(\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 \r…
\(\text { If } s_n=\sum_{r=0}^n T_r=\frac{(2 n-1)(2 n+1)(2 n+3)(2 n+5)}{64} \text { then find } \operatorname{Lim}_{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{T_r}=\) (22 Jan, Shift I, Memory Based)
\(\frac{2}{3}\)
\(\begin{aligned}& T_n=S_{n-} S_{n-1} \\& =\frac{(2 n-1)(2 n+1)(2 n+3)(2 n+5)-(2 n-3)(2 n-1)(2 n+1)(2 n+3)}{64} \\& T_n=\frac{(2 n-1)(2 n+1)(2 n+3)}{8} \\& \frac{1}{T_n}=\frac{8}{(2 n-1)(2 n+1)(2 n+3)} \\& \frac{1}{T_n}=2\left(\frac{1}{(2 n-1)(2 n+1)}-\frac{1}{(2 n-1)(2 n+3)}\right) \\& \sum_{r=1}^n \frac{1}{T_r}=2\left(\frac{1}{1 \times 3}-\frac{1}{(2 n-1)(2 n+3)}\right) \\& \operatorname{Lim}_{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{T_r}=\frac{2}{3}\end{aligned}\)
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