Question

$\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{r^3}{r^4+n^4}$ is equal to?

• A. $\log 2$
• B. $\frac{1}{4}\log 2$
• C. $\log \sqrt{2}$
• D. $e^2$

Answer 1

The correct option is B $\frac{1}{4}\log 2$

$\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{r^3}{r^4+n^4}$
$=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{\frac{1}{n}.{\left(\frac{r}{n}\right)}^3}{{\left(\frac{r}{n}\right)}^4+1}$
$={\int }_0^1\frac{x^3}{x^4+1}dx$

$($Lower limit $=\underset{n\to \infty }{lim}\frac{1}{n}=0$ and upper limit $\underset{n\to \infty }{lim}\frac{n}{n}=1)$
Put $x^4+1=t$
$\Rightarrow 4x^{3}dx=dt$
$I={\int }_1^2\frac{1}{4t}dt$
$\Rightarrow I=\frac{1}{4}[\log t{]}_1^2$
$I=\frac{1}{4}\log 2$