Question

Evaluate the sum: $$ \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} $$

• A. \( \frac{1}{4} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{1}{6} \)
• D. \( \frac{1}{3} \)

Hint:

Use partial fractions and telescoping series technique to evaluate complex infinite series.

Answer 1

The Correct Option is C

To evaluate the infinite sum \( \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} \), we employ partial fraction decomposition. First, express the summand as: \[ \frac{1}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2} \] Multiplying by the common denominator yields: \[ 1 = A(n+1)(n+2) + Bn(n+2) + Cn(n+1) \] Expanding and collecting terms: \[ 1 = (A + B + C)n^2 + (3A + 2B + C)n + 2A \] Equating coefficients: \[ A + B + C = 0 \] \[ 3A + 2B + C = 0 \] \[ 2A = 1 \] Solving these equations gives \(A = \frac{1}{2}\). Substituting this into the other equations leads to \(B = -\frac{1}{2}\) and \(C = 0\). The partial fraction decomposition is: \[ \frac{1}{n(n+1)(n+2)} = \frac{1}{2n} - \frac{1}{n+1} + \frac{1}{2(n+2)} \] Substituting this into the series: \[ \sum_{n=1}^{\infty} \left( \frac{1}{2n} - \frac{1}{n+1} + \frac{1}{2(n+2)} \right) \] This is a telescoping series. The sum evaluates to \( \frac{1}{6} \).