Question

The sum of infinite terms of the series $\frac{1}{\left(1+a\right)\left(2+a\right)}+\frac{1}{\left(2+a\right)\left(3+a\right)}+\frac{1}{\left(3+a\right)\left(4+a\right)}+....$where $a$ is a constant is

• A. $\frac{1}{1+a}$
• B. $\frac{2}{1+a}$
• C. infinity
• D. None of these

Answer 1

The correct option is A

$\frac{1}{1+a}$

Explanation for correct option:

The given series is $\frac{1}{\left(1+a\right)\left(2+a\right)}+\frac{1}{\left(2+a\right)\left(3+a\right)}+\frac{1}{\left(3+a\right)\left(4+a\right)}+....$

$$\begin{gathered} T_n=\sum _{n=1}^{\infty }\frac{1}{\left(n+a\right)\left(n+1+a\right)} \\ =\sum _{n=1}^{\infty }\left[\frac{1}{\left(n+a\right)}-\frac{1}{\left(n+1+a\right)}\right] \end{gathered}$$

Now,

$S_n=\sum _{}^{}{T}_n$

$$\begin{gathered} \Rightarrow S_n=\left[\frac{1}{\left(1+a\right)}-\frac{1}{\left(2+a\right)}\right]+\left[\frac{1}{\left(2+a\right)}-\frac{1}{\left(3+a\right)}\right]+...+\left[\frac{1}{\left(n+a\right)}-\frac{1}{\left(n+1+a\right)}\right] \\ \Rightarrow S_n=\frac{1}{\left(1+a\right)}-\frac{1}{\left(n+1+a\right)} \\ \Rightarrow \underset{n\to \infty }{\lim }{S}_n=\underset{n\to \infty }{\lim }\left[\frac{1}{\left(1+a\right)}-\frac{1}{\left(n+1+a\right)}\right] \\ \Rightarrow \underset{n\to \infty }{\lim }{S}_n=\frac{1}{\left(1+a\right)} \end{gathered}$$

Hence, option A is correct.