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

Solution to the Limit Calculation 

We are given that \( T_n = S_n - S_{n-1} \), where:

\[ T_n = \frac{1}{8} (2n-1)(2n+1)(2n+3) \] Now, let's simplify the expression for \( T_n \): \[ T_n = \frac{8}{(2n-1)(2n+1)(2n+3)} \]

Step 1: Setup the Sum

We are tasked with calculating the limit: \[ \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r} \] Substituting the expression for \( T_n \) into the sum: \[ \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r} = \lim_{n \to \infty} \frac{8}{4} \sum_{r=1}^n \frac{1}{(2n-1)(2n+1)(2n+3)} \] Simplifying the constant factor: \[ = 2 \sum_{r=1}^n \frac{1}{(2r-1)(2r+1)(2r+3)} \]

Step 2: Breaking the Series

The series can be expressed as a telescoping series. Observing the pattern in the terms: \[ \sum_{r=1}^n \frac{1}{(2r-1)(2r+1)(2r+3)} = \left( \frac{1}{1 \cdot 3} - \frac{1}{3 \cdot 5} \right) + \left( \frac{1}{3 \cdot 5} - \frac{1}{5 \cdot 7} \right) + \dots \] This is a telescoping series, where many terms cancel out, leaving us with: \[ \lim_{n \to \infty} 2 \left( \frac{1}{1 \cdot 3} \right) \]

Step 3: Final Simplification

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

Conclusion

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

Answer 2

1. Write \(T_r\) as the difference: \[ T_r = S_r - S_{r-1},\qquad S_n=\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}. \] Compute (exactly) \[ T_r = S_r - S_{r-1}=\frac{(2r-1)(2r+1)(2r+3)}{8}. \]
2. So \[ \frac{1}{T_r}=\frac{8}{(2r-1)(2r+1)(2r+3)}. \] Perform partial fractions: \[ \frac{8}{(2r-1)(2r+1)(2r+3)} =\frac{1}{2r-1}-\frac{2}{2r+1}+\frac{1}{2r+3}. \]
3. The sum telescopes: \[ \sum_{r=1}^{n}\frac{1}{T_r} =\sum_{r=1}^{n}\Big(\frac{1}{2r-1}-\frac{2}{2r+1}+\frac{1}{2r+3}\Big). \] Summing and simplifying gives the closed form \[ \sum_{r=1}^{n}\frac{1}{T_r}=\frac{8n(n+2)}{3(4n^2+8n+3)}. \]
4. Take the limit as \(n\to\infty\). The leading terms give \[ \lim_{n\to\infty}\frac{8n(n+2)}{3(4n^2+8n+3)} =\frac{8}{3\cdot 4}=\frac{2}{3}. \]

Answer 

\(\displaystyle \lim_{n\to\infty}\sum_{r=1}^{n}\frac{1}{T_r}=\frac{2}{3}.\) (Option 3)