Question

If \[ \sum_{r=1}^{n} a_r = \frac{n(n+1)(n+2)}{6}\quad \forall\, n \ge 1, \] then \[ \lim_{n\to\infty}\sum_{r=1}^{n}\frac{1}{a_r} = \]

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

Hint:

When a sum \(S_n\) is given explicitly, always find \[ a_r = S_r - S_{r-1} \] Many series of the form \(\dfrac{1}{r(r+1)}\) lead to telescoping sums.

Answer 1

The Correct Option is C

Step 1: Express \(a_r\) using partial sums. Given: \[ S_n = \sum_{r=1}^{n} a_r = \frac{n(n+1)(n+2)}{6} \] Then, \[ a_r = S_r - S_{r-1} \]
Step 2: Find \(a_r\). \[ S_r = \frac{r(r+1)(r+2)}{6}, \quad S_{r-1} = \frac{(r-1)r(r+1)}{6} \] \[ a_r = \frac{r(r+1)}{6}\big[(r+2)-(r-1)\big] \] \[ a_r = \frac{r(r+1)}{6}\times 3 = \frac{r(r+1)}{2} \]
Step 3: Write the required sum. \[ \sum_{r=1}^{n}\frac{1}{a_r} = \sum_{r=1}^{n}\frac{2}{r(r+1)} \]
Step 4: Use partial fractions. \[ \frac{2}{r(r+1)} = 2\left(\frac{1}{r}-\frac{1}{r+1}\right) \] \[ \sum_{r=1}^{n}\frac{1}{a_r} = 2\sum_{r=1}^{n}\left(\frac{1}{r}-\frac{1}{r+1}\right) \]
Step 5: Evaluate the telescoping sum. \[ =2\left(1-\frac{1}{n+1}\right) \]
Step 6: Take the limit as \(n\to\infty\). \[ \lim_{n\to\infty}2\left(1-\frac{1}{n+1}\right)=2 \]
Hence, \[ \boxed{2} \]