Question

If \(\sum\)\(_{r=1}^n T_r\) = \(\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}\) , then \( \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r} \) is equal to :

• A. 1
• B. 0
• C. \( \frac{2}{3} \)
• D. \( \frac{1}{3} \)

Hint:

Use the limit properties of series to find the summation at infinity.

Answer 1

The Correct Option is C

We are given that \( T_n = S_n - S_{n-1} \). Thus: \[ T_n = \frac{1}{8} (2n-1)(2n+1)(2n+3) \] Simplifying further: \[ T_n = \frac{8}{(2n-1)(2n+1)(2n+3)} \] Now, calculate the limit: \[ \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r} = \lim_{n \to \infty} \frac{8}{4} \sum_{r=1}^n \frac{1}{(2n-1)(2n+1)(2n+3)} \] \[ = \frac{8}{4} \left( \frac{1}{1.3} + \frac{1}{1.5} + \cdots \right) \] \[ = \lim_{n \to \infty} 2 \left[ \left( \frac{1}{1 \cdot 3} - \frac{1}{3 \cdot 5} \right) + \left( \frac{1}{3 \cdot 5} - \frac{1}{5 \cdot 7} \right) + \dots \right] \] Thus, the answer is: \[ \frac{2}{3} \]