Question

Let $S=\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+\cdots \text{upto}\infty$. Then the value of $2S$ is

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

Answer 1

The correct option is B $\frac{1}{2}$

$S=\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+\cdots \text{upto}\infty$

$\Rightarrow S=\sum _{n=2}^{\infty }\frac{1}{n^3-n}$

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

Let $u_n=\frac{1}{(n-1)n}$

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

$\Rightarrow S=\frac{1}{2}(u_2-\underset{n\to \infty }{lim}\frac{1}{n(n+1)})$

$\Rightarrow S=\frac{1}{2}(\frac{1}{2}-0)$

$\Rightarrow S=\frac{1}{4}$