Question

The value of $\frac{1}{3}+\frac{1}{8}+\frac{1}{15}+\frac{1}{24}+\cdots +\infty$ is

• A. $\frac{1}{2}$
• B. $\frac{3}{4}$
• C. $\frac{3}{2}$
• D. $\frac{3}{5}$

Answer 1

The correct option is B $\frac{3}{4}$

$S=\frac{1}{3}+\frac{1}{8}+\frac{1}{15}+\frac{1}{24}+\cdots +\infty =\sum _{n=2}^{\infty }\frac{1}{n^2-1}$
$\Rightarrow S=\sum _{n=2}^{\infty }\frac{1}{(n+1)(n-1)}$

$\Rightarrow S=\frac{1}{2}\sum _{n=2}^{\infty }\frac{(n+1)-(n-1)}{(n+1)(n-1)}$
$\Rightarrow 2S=\sum _{n=2}^{\infty }\frac{1}{(n-1)}-\frac{1}{(n+1)}$

$=+\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+\frac{1}{5}-\frac{1}{7}...+\frac{1}{n-1}-\frac{1}{n+1}$

$\Rightarrow 2S=\frac{1}{1}+\frac{1}{2}-\frac{1}{n+1}=\frac{3}{2}-\frac{1}{n+1}$
As $n\to \infty ,2S\to \frac{3}{2}$
$\therefore S=\frac{3}{4}$