$\lim _{n \rightarrow \infty} \frac{1}{n}\left\{\sin ^5\left(\frac{\pi}{6 n}\right)+\sin ^5\left(\frac{2…
$\lim _{n \rightarrow \infty} \frac{1}{n}\left\{\sin ^5\left(\frac{\pi}{6 n}\right)+\sin ^5\left(\frac{2 \pi}{6 n}\right)+\sin ^5\left(\frac{3 \pi}{6 n}\right)+\right.$
$\left.+\sin ^5\left(\frac{\pi}{2}\right)\right\}=$
- $\frac{8}{15 \pi}$
- $\frac{8}{5 \pi}$
- $\frac{32}{5 \pi}$
- $\frac{16}{5 \pi}$
Solution
$\begin{aligned} \lim _{n \rightarrow \infty} \frac{1}{n}\left\{\sin ^5\left(\frac{\pi}{6 n}\right)+\sin ^5\left(\frac{2 \pi}{6 n}\right)\right. & +\sin ^5\left(\frac{3 \pi}{6 n}\right) \\ & \left.+\ldots+\sin ^5\left(\frac{\pi}{2}\right)\right\}\end{aligned}$
$
\begin{aligned}
=\lim _{n \rightarrow \infty} \frac{1}{n}\left\{\sin ^5\left(\frac{\pi}{6 n}\right)+\sin ^5\left(\frac{2 \pi}{6 n}\right)\right. & +\sin ^5\left(\frac{3 \pi}{6 n}\right) \\
& \left.+\ldots+\sin ^5\left(\frac{3 n \pi}{6 n}\right)\right\} \\
= & \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{3 n} \sin ^5\left(\frac{r \pi}{6 n}\right)=\int_0^3 \sin ^5\left(\frac{\pi}{6} x\right) d x
\end{aligned}
$
Let
$
\frac{\pi}{6} x=t
$
For upper limit at $x=3, t=\pi / 2$ and lower limit at $x=0, t=0$ and $d x=\frac{6}{\pi} d t$
So,
$
\begin{aligned}
\int_0^3 \sin ^5\left(\frac{\pi}{6} x\right) d t & =\frac{6}{\pi} \int_0^{\pi / 2} \sin ^5(t) d t \\
& =\frac{6}{\pi} \times \frac{4 \times 2}{5 \times 3 \times 1}=\frac{16}{5 \pi}
\end{aligned}
$
Asked in: AP EAMCET 2018 (23 Apr Shift 2)
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