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$$\lim _{n\to \infty }\frac{1}{n^3}\sum _{k=1}^{2n}\frac{k^2}{8}\arcsin \left(\frac{k}{2n}\right) =\: L$$

I kind of know how to solve a Riemann sum with $n$ terms. I create a division, usually $D_n=(0, \frac{1}{n},..., \frac{n}{n})$, with some intermediary points (usually $\epsilon_k = \frac{k}{n}$, with $k$ from $0$ to $n-1$ or from $1$ to $n$) and then I can say that the limit of the series is $\int _0^1 f(x)$, after $k/n = x$. However, how can i do that form $2n$ terms ? I thought about rewriting the limit as this $$L=\lim _{n\to \infty }\frac{1}{2n}\sum _{k=1}^{2n}\:\frac{1}{4}\left(\frac{k}{n}\right)^2\arcsin \left(\frac{k}{n}\cdot \frac{1}{2}\right)$$ and then form the function by $\frac{k}{n} = x$. The correct answer should be $\frac{\pi }{6}-\frac{2}{9}$.

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    $\begingroup$ Just have the same, with $\frac{k}{2n}$ instead of $\frac{k}{n}$. If that helps conceptually, set $m=2n$ and rewrite the sum as $\frac{8}{m^3}\sum_{k=1}^m \frac{k^2}{8}\arcsin\frac{k}{m}$. $\endgroup$ – Clement C. Feb 25 '17 at 20:55
  • $\begingroup$ Related : math.stackexchange.com/questions/469885/… $\endgroup$ – lab bhattacharjee Feb 26 '17 at 5:07
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Just do exactly the same, with $\frac{k}{2n}$ instead of $\frac{k}{n}$. If that helps conceptually, set $m=2n$ and rewrite the sum as $$\frac{1}{n^3}\sum_{k=1}^{2n} \frac{k^2}{8}\arcsin\frac{k}{2n}=\frac{8}{m^3}\sum_{k=1}^m \frac{k^2}{8}\arcsin\frac{k}{m}$$ Then, $$ \frac{8}{m^3}\sum_{k=1}^m \frac{k^2}{8}\arcsin\frac{k}{m} = \frac{1}{m}\sum_{k=1}^m \left(\frac{k}{m}\right)^2\arcsin\frac{k}{m} $$ and now you should recognize a Riemann sum converging to $\int_0^1 x^2\arcsin x dx$ when $m\to\infty$ (or, equivalently, $n\to\infty$).

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Since $x^2\arcsin(x)$ is a Riemann-integrable function over $(0,1)$,

$$ \lim_{n\to +\infty}\frac{1}{2n}\sum_{k=1}^{2n}\left(\frac{k}{2n}\right)^2 \arcsin\left(\frac{k}{2n}\right) = \int_{0}^{1} x^2\arcsin(x)\,dx $$ and the last integral can be computed through the substitution $x=\sin\theta$ and integration by parts.

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