Question

The value of $\sum _{n=1}^{\infty }\frac{3n^2+3n+1}{{\left(n^2+n\right)}^3}$ equals to

• A. $\frac{1}{4}$
• B. $\frac{1}{8}$
• C. $\frac{1}{2}$
• D. 1

Answer 1

Answer: (D)

$\sum _{n=1}^{\infty }\frac{3n^2+3n+1}{n^3(n+1{)}^3}=\sum _{n=1}^{\infty }\left(\frac{1}{n^3}-\frac{1}{(n+1{)}^3}\right)=1$