Question

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

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

Answer 1

Answer: (A)

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

The general term of the above expression is
$\frac{1}{(n+a)(n+1+a)}$
$=\frac{n+1+a-(n+a)}{(n+a)(n+1+a)}$
$=\frac{1}{n+a}-\frac{1}{n+1+a}$
Hence the series becomes
$=\frac{1}{1+a}-\frac{1}{a+2}+\frac{1}{2+a}-\frac{1}{3+a}+....\infty$
$=\frac{1}{1+a}$