Question

The sum of the series \[ \frac{1}{2(2^2 - 1)} + \frac{1}{3(3^2 - 1)} + \frac{1}{4(4^2 - 1)} + \cdots \] is ...........

Hint:

Whenever a rational term involves \( n(n-1)(n+1) \), use partial fractions — it usually telescopes.

Answer 1

Correct Answer: 0.25

Step 1: Express the general term.
The \( n^{th} \) term is \[ T_n = \frac{1}{n(n^2 - 1)} = \frac{1}{n(n-1)(n+1)}. \]

Step 2: Partial fraction decomposition.
\[ \frac{1}{n(n-1)(n+1)} = \frac{A}{n-1} + \frac{B}{n} + \frac{C}{n+1}. \] Simplifying gives \( A = \frac{1}{2}, \, B = -1, \, C = \frac{1}{2}. \) Thus, \[ T_n = \frac{1}{2(n-1)} - \frac{1}{n} + \frac{1}{2(n+1)}. \]

Step 3: Write as telescoping series.
\[ S_N = \sum_{n=2}^{N} T_n = \frac{1}{2}\left( \frac{1}{1} - \frac{1}{2} \right) + \frac{1}{2}\left( \frac{1}{2} - \frac{1}{3} \right) + \cdots \] On simplification, most terms cancel out.

Step 4: Limit as \( N \to \infty. \)
The sum converges to \[ S = \frac{1}{4}. \]

Final Answer: \[ \boxed{\frac{1}{4}} \]