Question

If the sum of infinite terms of the series $\frac{1}{3.4}+\frac{1}{4.5}+...$ is $L$, then value of $6L$ is

Answer 1

Sum of $n$ terms of the series $\frac{1}{3\times 4}+\frac{1}{4\times 5}+...+\frac{1}{(n+2)(n+3)}$
$=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{n+2}-\frac{1}{n+3}$
$=\frac{1}{3}-\frac{1}{n+3}$
For infinite terms, $n$ tends to infinity and sum of the series approaches $\frac{1}{3}$
Hence, $L=\frac{1}{3}$
$6L=2$