JEE Main 2025 — 22 January, Morning Shift — Question 10

$\mathrm{T}_{\mathrm{n}}=\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1}$

$\Rightarrow \mathrm{T}_{\mathrm{n}}=\frac{1}{8}(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3)$

$\Rightarrow \frac{1}{\mathrm{~T}_{\mathrm{n}}}=\frac{8}{(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3)}$

$\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{T_{r}}=\lim _{n \rightarrow \infty} 8 \sum_{r=1}^{n} \frac{1}{(2 n-1)(2 n+1)(2 n+3)}$

$=\lim _{n \rightarrow \infty} \frac{8}{4} \sum\left(\frac{1}{(2 n-1)(2 n+1)}-\frac{1}{(2 n+1)(2 n+3)}\right)$

$=\lim _{n \rightarrow \infty} 2\left[\left(\frac{1}{1.3}-\frac{1}{3.5}\right)+\left(\frac{1}{3.5}-\frac{1}{5.7}\right)+\ldots\right]$

$=\frac{2}{3}$.