Calculate $\lim _{n\to \infty }a_n\int _0^1 x^{2n}\sin \frac{\pi x}{2}dx$ I have to calculate
$$\lim _{n\to \infty }a_n\int _0^1 x^{2n}\sin \frac{\pi x}{2}dx$$
Where $$a_n = \sum _{k=1}^n\sin \frac{k\pi }{2n}$$
I have found that $$\lim _{n\to \infty} \frac{a_n}{n} = \frac{2}{\pi} $$ if that helps in any way.
 A: Hint. One may use an integration by parts, for $n\ge1$, 
$$
\begin{align}
\int_0^1 x^{2n}\sin \frac{\pi x}{2}\:dx&=\left[\frac{x^{2n+1}}{2n+1}\cdot \sin \frac{\pi x}{2}\right]_0^1-\frac{\pi}{2(2n+1)}\int_0^1 x^{2n+1}\cos \frac{\pi x}{2}\:dx
\\\\&=\frac{1}{2n+1}-\frac{\pi}{2(2n+1)}\int_0^1 x^{2n+1}\cos \frac{\pi x}{2}\:dx.
\end{align}
$$ Then observing that, as $n \to \infty$,
$$
\left|\int_0^1 x^{2n+1}\cos \frac{\pi x}{2}\:dx\right|\le\int_0^1 \left|x^{2n+1}\right|\:dx=\frac{1}{2n+2} \to 0,
$$ one obtains, as $n \to \infty$,
$$
n\int_0^1 x^{2n}\sin \frac{\pi x}{2}\:dx=\frac{n}{2n+1}-\frac{\pi\cdot n}{2(2n+1)}\int_0^1 x^{2n+1}\cos \frac{\pi x}{2}\:dx \to \frac12.
$$ By writing, as $n \to \infty$,
$$
a_n\int _0^1 x^{2n}\sin \frac{\pi x}{2}dx=\color{blue}{\frac{a_n}n} \cdot n\int _0^1 x^{2n}\sin \frac{\pi x}{2}dx
$$one deduces an answer to the initial question.
A: Hint: A problem that has been here many times: If $f$ is continuous on $[0,1],$ then
$$\lim_{n \to \infty}(n+1)\int_0^1 x^n f(x)\,dx = f(1).$$
Use this with $f(x) = \sin (\pi x /2)$ and your result for $a_n.$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
\mbox{With Laplace Method,}\
\int_{0}^{1}x^{2n}\sin\pars{{\pi \over 2}\, x}\,\dd x =
\int_{0}^{1}\pars{1 - x}^{2n}\cos\pars{{\pi \over 2}\, x}\,\dd x\sim {1 \over 2n}\
\mbox{as}\ n \to \infty
\end{align}

So, you are left with

\begin{align}
\lim_{n \to \infty}\pars{a_{n}\,{1 \over 2n}} = \bbx{\ds{1 \over \pi}}
\end{align}
