Question

Value of $\sum_{n=0}^{\infty}\frac{2}{(2n+1)(2n+3)}$ is

• A. $\infty$
• B. 1
• C. $\frac{1}{3}$
• D. $\frac{4}{6}$

Hint:

For infinite telescoping series, the sum is simply the value of the "uncancelled" terms at the beginning of the sequence.

Answer 1

The Correct Option is B

This problem requires the use of partial fraction decomposition to find the sum of an infinite series.
Step 1: Let the general term be $a_n = \frac{2}{(2n+1)(2n+3)}$. We can write this as: $\frac{2}{(2n+1)(2n+3)} = \frac{A}{2n+1} + \frac{B}{2n+3}$. Solving for constants: $2 = A(2n+3) + B(2n+1)$. Set $2n+1 = 0 \implies n = -1/2$, then $2 = A(2) \implies A = 1$. Set $2n+3 = 0 \implies n = -3/2$, then $2 = B(-2) \implies B = -1$. So, $a_n = \frac{1}{2n+1} - \frac{1}{2n+3}$.
Step 2: The sum for the first $k$ terms is: $S_k = \sum_{n=0}^{k} \left( \frac{1}{2n+1} - \frac{1}{2n+3} \right)$ $S_k = (1 - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{7}) + \dots + (\frac{1}{2k+1} - \frac{1}{2k+3})$.
Step 3: Observing the sequence, we see that $-1/3$ cancels with $+1/3$, $-1/5$ with $+1/5$, and so on. The only terms remaining after full cancellation are the first part of the first term and the last part of the last term: $S_k = 1 - \frac{1}{2k+3}$.
Step 4: To find the sum of the infinite series, we take the limit as $k \to \infty$: $S = \lim_{k \to \infty} \left( 1 - \frac{1}{2k+3} \right)$. As $k$ becomes infinitely large, the fraction $\frac{1}{2k+3}$ approaches zero. $S = 1 - 0 = 1$. The result is 1, which corresponds to Option (2).