Question

Evaluate : $\underset{n\to \infty }{\lim }\left[\frac{1}{n^3+1}+\frac{4}{n^3+1}+\frac{9}{n^3+1}+\dots \dots +\frac{n^2}{n^3+1}\right]$

Answer 1

Answer: 0.33

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