Question

The value of $\underset{n\to \infty }{lim}\frac{1}{n^2}\{{\sin }^2\frac{\pi }{4n}+2{\sin }^2\frac{2\pi }{4n}+\cdots +n{\sin }^2\frac{4\pi }{4n}\}$ is

• A. $\frac{1}{2}-\frac{2}{\pi }+\frac{4}{{\pi }^2}$
• B. $\frac{1}{4}-\frac{1}{\pi }+\frac{2}{{\pi }^2}$
• C. $\frac{1}{4}+\frac{1}{\pi }-\frac{2}{{\pi }^2}$
• D. $\frac{1}{2}-\frac{2}{\pi }$

Answer 1

The correct option is B $\frac{1}{4}-\frac{1}{\pi }+\frac{2}{{\pi }^2}$

Reimann sum
$\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=0}^{n}f\left(\frac{r}{n}\right)={\int }_0^{1}f\left(x\right)dx.$
$\underset{n\to \infty }{lim}\frac{1}{n^2}\{{\sin }^2\frac{\pi }{4n}+2{\sin }^2\frac{2\pi }{4n}+\cdots +n{\sin }^2\frac{n\pi }{4n}\}$
$=\underset{n\to \infty }{lim}\frac{1}{n}\{\frac{1}{n}{\sin }^2\frac{\pi }{4n}+\frac{2}{n}{\sin }^2\frac{2\pi }{4n}+\frac{3}{n}{\sin }^2\frac{3\pi }{4n}+\cdots +\frac{n}{n}{\sin }^2\frac{n\pi }{4n}\}$
$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n(\frac{r}{n}{\sin }^2\frac{\pi }{4n}\times r+0)$
$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=0}^n\left(\frac{r}{n}\right){\sin }^2(\frac{\pi }{4}\times \left(\frac{r}{n}\right))$
$={\int }_0^{1}x{\sin }^2\frac{\pi x}{4}dx={\int }_0^{1}x\left(\frac{1-\cos \frac{\pi x}{2}}{2}\right)dx$
$=\frac{1}{2}[{\frac{x^2}{2}|}_0^1-{\int }_0^{1}x\cos \left(\frac{\pi x}{2}\right)]$
$=\frac{1}{2}[{\frac{x^2}{2}|}_0^1-\{{x\frac{\sin \frac{\pi x}{2}}{\frac{\pi }{2}}|}_0^1-{\int }_0^1\frac{\sin \frac{\pi x}{2}}{\frac{\pi }{2}}dx\}]=\frac{1}{4}-\frac{1}{\pi }+\frac{2}{{\pi }^2}$