If $\sum_{r=1}^{n} T_{r} = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$, then $\lim_{n \to \infty} \sum_{r=1}^{n} \left(\frac{1}{T_{r}}\right)$ is equal to:
- (1) 1
- (2) 0
- (3) $\frac{2}{3}$
- (4) $\frac{1}{3}$
Solution:
Let $S_n = \sum_{r=1}^{n} T_r$. The $n^{th}$ term $T_n$ is given by $S_n – S_{n-1}$.
Based on the structure, $T_n = \frac{1}{8}(2n-1)(2n+1)(2n+3)$.
We need to find $\lim_{n \to \infty} \sum \frac{1}{T_n}$:
$$ \frac{1}{T_n} = \frac{8}{(2n-1)(2n+1)(2n+3)} $$
Using partial fractions (telescoping series):
$$ \frac{1}{(2n-1)(2n+1)(2n+3)} = \frac{1}{4} \left[ \frac{1}{(2n-1)(2n+1)} – \frac{1}{(2n+1)(2n+3)} \right] $$
So, sum $S = 8 \times \frac{1}{4} \sum_{n=1}^{\infty} \left[ \frac{1}{(2n-1)(2n+1)} – \frac{1}{(2n+1)(2n+3)} \right]$.
$$ S = 2 \left[ \left(\frac{1}{1 \cdot 3} – \frac{1}{3 \cdot 5}\right) + \left(\frac{1}{3 \cdot 5} – \frac{1}{5 \cdot 7}\right) + \dots \right] $$
This is a telescoping sum. The terms cancel out, leaving only the first term.
$$ S = 2 \left[ \frac{1}{3} – 0 \right] = \frac{2}{3} $$
Ans. (3)