If , then is equal to, Sequence and series, Sum to n terms of Special Series, Q no 5 [2025]

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Solution

Q.
If $\sum_{r = 1}^{n} T_{r} = \frac{(2 n - 1) (2 n + 1) (2 n + 3) (2 n + 5)}{64}$ , then $\lim_{n \to \infty} \sum_{r = 1}^{n} (\frac{1}{T_{r}})$ is equal to : [2025]

1 0

2 $\frac{2}{3}$

3 $\frac{1}{3}$

4 1

Ans.
(2)

Given, $S_{n} = \sum_{r = 1}^{n} T_{r} = \frac{(2 n - 1) (2 n + 1) (2 n + 3) (2 n + 5)}{64}$

So, $S_{n - 1} = \frac{(2 n - 3) (2 n - 1) (2 n + 1) (2 n + 3)}{64}$

$∵ T_{n} = S_{n} - S_{n - 1}$

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

$∴ \sum_{r = 1}^{n} (\frac{1}{T_{n}}) = \frac{8}{4} [\frac{1}{(2 n - 1) (2 n + 1)} - \frac{1}{(2 n + 1) (2 n + 3)}]$

$\lim_{n \to \infty} \sum_{r = 1}^{n} (\frac{1}{T_{r}}) = 2 \lim_{n \to \infty} (\frac{1}{(2 n - 1) (2 n + 1)} - \frac{1}{(2 n + 1) (2 n + 3)})$

$\Rightarrow \lim_{n \to \infty} \sum_{r = 1}^{n} (\frac{1}{T_{r}}) = 2 (\frac{1}{1 \cdot 3} - \frac{1}{3 \cdot 5} + \frac{1}{3 \cdot 5} - \frac{1}{5 \cdot 7} + . . .)$

$\lim_{n \to \infty} \frac{2}{3} - \frac{2}{(2 n + 1) (2 n + 3)} = \frac{2}{3}$ .

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