Question

$\underset{x\to \infty }{lim}(\frac{1}{n^3+1}+\frac{4}{n^3+1}+\frac{9}{n^3+1}+\dots \dots +\frac{n^2}{n^3+1})$ is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$\underset{n\to \infty }{lim}\left(\frac{1+4+9+\dots +n^2}{n^3+1}\right)=\underset{n\to \infty }{lim}\left(\frac{\Sigma n^2}{n^3+1}\right)=\underset{n\to \infty }{lim}\frac{n(n+1)(2n+1)}{6(n^3+1)}=\underset{n\to \infty }{lim}\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6(1+\frac{1}{n^3})}=\frac{1\times 2}{6\times 1}=\frac{1}{3}$