JEE Math Practice Question
If $\sum\limits_{r=1}^n T_r=\frac{(2 n-1)(2 n+1)(2 n+3)(2 n+5)}{64}$, then $\lim \limits_{n \rightarrow \infty} \sum\limits_{r=1}^n\left(\frac{1}{T_r}\right)$ is equal to :
- A.$\frac{2}{3}$
- B.$\frac{1}{3}$
- C.1
- D.0
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