Question

Sum to infinite terms of the series $\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{(n+2)(n+3)}$ is

• A. $\frac{1}{6}$
• B. $\frac{1}{3}$
• C. $\frac{1}{12}$
• D. $\frac{n}{3(n+3)}$

Answer 1

The correct option is B $\frac{1}{3}$

Sum $=\frac{1}{3.4}+\frac{1}{4.5}+................+\frac{1}{(n+2)(n+3)}$
$=\frac{4-3}{3.4}+\frac{5-4}{4.5}+................+\frac{(n+3)-(n+2)}{(n+2)(n+3)}$
$=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+................+\frac{1}{(n+2)}-\frac{1}{(n+3)}$
$=\frac{1}{3}-\frac{1}{(n+3)}$
given $n\to \infty \Rightarrow sum=\frac{1}{3}$