TITLE: Calculate $\lim _{n\to \infty }a_n\int _0^1 x^{2n}\sin \frac{\pi x}{2}dx$
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}