Question

$n\stackrel{Lt}{\to }\infty [\frac{1}{1+n^3}+\frac{4}{8+n^3}+\frac{9}{27+n^3}+\dots .+\frac{1}{2n}]=$

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

Answer 1

Answer: (C)

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

$n\stackrel{Lt}{\to }\infty [\frac{1}{(1+n^3)}+\frac{4}{(8+n^3)}+\frac{9}{(27+n^3)}+.......\frac{n^2}{(n^3+n^3)}]$
$n\stackrel{Lt}{\to }\infty {\sum }_{r=1}^n\frac{r^2}{r^3+n^3}=n\stackrel{Lt}{\to }\infty {\sum }_{r=1}^n\frac{1}{n}\frac{(\frac{r}{n}{)}^2}{(\frac{r}{n}{)}^3+1}$
$={\int }_0^1\frac{x^2}{x^3+1}dx=\frac{1}{3}{\int }_1^2\frac{d{x}^3}{1+x^3}$
$=\frac{1}{3}log(1+x^3){\int }_0^1$
$=\frac{1}{3}\left[log\right(2\left)\right]$
$=\frac{1}{3}log2$