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

Calculation of a Limit

Given \( T_n = S_n - S_{n-1} \) where \( T_n = \frac{1}{8} (2n-1)(2n+1)(2n+3) \). The expression for \( T_n \) is simplified as: \[ T_n = \frac{8}{(2n-1)(2n+1)(2n+3)} \]

Step 1: Formulating the Sum

The objective is to compute the limit: \[ \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r} \] Substituting the expression for \( T_n \): \[ \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r} = \lim_{n \to \infty} \frac{8}{4} \sum_{r=1}^n \frac{1}{(2n-1)(2n+1)(2n+3)} \] The constant factor is simplified: \[ = 2 \sum_{r=1}^n \frac{1}{(2r-1)(2r+1)(2r+3)} \]

Step 2: Decomposing the Series

The series is a telescoping series. Observing the pattern of the terms: \[ \sum_{r=1}^n \frac{1}{(2r-1)(2r+1)(2r+3)} = \left( \frac{1}{1 \cdot 3} - \frac{1}{3 \cdot 5} \right) + \left( \frac{1}{3 \cdot 5} - \frac{1}{5 \cdot 7} \right) + \dots \] Due to cancellations in this telescoping series, the sum simplifies to: \[ \lim_{n \to \infty} 2 \left( \frac{1}{1 \cdot 3} \right) \]

Step 3: Final Calculation

The limit of the series as \( n \to \infty \) is: \[ \frac{2}{3} \]

Result

The value of the limit is: \[ \boxed{\frac{2}{3}} \]