Question

Let $f\left(n\right)=\frac{1}{n(n^2-1)}$. The series ${\sum }_{n=1}^{\infty }f(n+1)$ is convergent to the sum?

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

Answer 1

The correct option is B $\frac{1}{4}$

$f\left(n\right)=\frac{1}{n(n^2-1)}=\frac{1}{2}(\frac{1}{n(n-1)}-\frac{1}{n(n+1)})$

$f(n+1)=\frac{1}{2}(\frac{1}{(n+1)n}-\frac{1}{(n+1)(n+2)})$

${\Sigma }_{n=1}^{n=\infty }f(n+1)=\frac{1}{2}(\frac{1}{1\times 2}-\frac{1}{2\times 3}+\frac{1}{2\times 3}-\frac{1}{3\times 4}+\frac{1}{3\times 4}\cdots \cdots )$

${\Sigma }_{n=1}^{n=\infty }f(n+1)=\frac{1}{4}$

Correct option is B