Question

$\underset{n\to \infty }{lim}[\frac{1^2}{1^3+n^3}+\frac{2^2}{2^3+n^3}+...+\frac{1}{2n}]$ is equal to

• A. $\frac{1}{3}\log 2$
• B. $\frac{1}{2}\log 2$
• C. $\log 2$
• D. none of these

Answer 1

Answer: (A)

$\frac{1}{3}\log 2$

$\underset{n\to \infty }{lim}[\frac{1^2}{1^3+n^3}+\frac{2^2}{2^3+n^3}+...+\frac{1}{2n}]$

$=\underset{n\to \infty }{lim}\sum _{r=1}^n\left(\frac{r^2}{r^3+n^3}\right)=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n\frac{{\left(\frac{r}{n}\right)}^2}{{\left(\frac{r}{n}\right)}^3+1}$

$={\int }_0^1\frac{x^2}{x^3+1}dx$

$=\frac{1}{3}{\left[\log (x^3+1)\right]}_0^1$

$=\frac{1}{3}(\log 2-\log 1)$

$=\frac{1}{3}\log 2$