Evaluating $\lim_{n\to\infty}\sum_{k=1}^n \frac{\pi k}{2n}\int_0^1 x^{2n}\sin\frac{\pi x}{2}dx$

Question

Hello I am trying to compute $$\lim_{n\to\infty}\sum_{k=1}^n \frac{\pi k}{2n}\int_0^1 x^{2n}\sin\frac{\pi x}{2}dx$$ I have rewrited it as: $$\lim_{n\to\infty}\frac1n\sum_{k=1}^n \frac{k}{n}\int_0^1\frac{n\pi}{2} x^{2n}\sin\frac{\pi x}{2}dx$$ Now the first part is just a Riemann integral $\int_0^1 x \,dx=\frac{1}{2}\,$ however I dont know if I am allowed to use it in this case. Now for the second part using $\frac{\pi x}{2}=t$ we have:$$\int_0^1\frac{n\pi}{2} x^{2n}\sin\frac{\pi x}{2}dx=\left(\frac{\pi}{2}\right)^nn\int_0^\frac{\pi}{2}t^{2n}\sin t\,dt=\left(\frac{\pi}{2}\right)^n\frac{n}{2n+1}\int_0^\frac{\pi}{2}\left(t^{2n+1}\right)'\sin t\, dt$$ So integrating by parts two times gives: $$I(2n)=\frac{1}{2n+1}\left(\left(\frac{\pi}{2}\right)^{2n+1}-\frac{1}{2n+2}I(2n+2)\right)$$ or rewritten as:$$I(2n)=2n\left(\left(\frac{\pi}{2}\right)^{2n-1}-(2n-1)I(2n-2)\right)$$ But how do I use the relation? Could you give me some help with this?

EDIT:Originally was:$$\lim_{n\to\infty}\sum_{k=1}^n \sin \frac{\pi k}{2n}\int_0^1 x^{2n}\sin\frac{\pi x}{2}dx$$

Answer 1

You were on the right track using integration by parts twice. Here, we present an approach that relies only on elementary tools. To that end we proceed.

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Using $\sum_{k=1}^n k=\frac{n(n+1)}{2}$, we see that

$$\sum_{k=1}^n \frac{\pi k}{2n}=\frac{(n+1)\pi}{4}$$

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Next, integrating by parts with $u=\sin(\pi x/2)$ and $v=\frac{x^{2n+1}}{2n+1}$ reveals

$$\int_0^1 x^{2n}\sin(\pi x/2)\,dx=\frac1{2n+1}-\frac{\pi}{2(2n+1)}\int_0^1 x^{2n+1}\cos(\pi x/2)\,dx\tag1$$

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Subsequently, integrating by parts the integral on the right-hand side of $(1)$ with $u=\cos(\pi x/2)$ and $v=\frac{x^{2n+2}}{2n+2}$ gives

$$\int_0^1 x^{2n}\sin(\pi x/2)\,dx=\frac1{2n+1}-\frac{\pi}{2(2n+1)}\frac{\pi }{2(2n+2)}\int_0^1 x^{2n+2}\sin(\pi x/2)\,dx\tag2$$

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Multiplying $(2)$ by $\frac{(n+1)\pi}{4}$ and letting $n\to \infty$ yields the coveted limit

$$\lim_{n\to \infty}\left(\sum_{k=1}^n\frac{\pi k}{2n}\int_0^1 x^{2n}\sin(\pi x/2)\,dx\right)=\frac\pi8$$

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Tools Used: Integration by Parts and the sum $\displaystyle \sum_{k=1}^n k=\frac{n(n+1)}{2}$

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INTENDED LIMIT:

In a comment, the OP stated that the original problem was to evaluate $$\lim_{n\to \infty}\left(\sum_{k=1}^n\sin\left(\frac{\pi k}{2n}\right)\int_0^1 x^{2n}\sin(\pi x/2)\,dx\right)$$

We proceed herein to evaluate the intended limit.

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We note that $\sum_{k=1}^n \sin\left(\frac{\pi k}{2n}\right)=\frac12 \cot\left(\frac{\pi}{4n}\right)+\frac12$. Using $(2)$, we see that

$$\begin{align} \sum_{k=1}^n\sin\left(\frac{\pi k}{2n}\right)\int_0^1 x^{2n}\sin(\pi x/2)\,dx&=\frac12\left(\cot\left(\frac{\pi}{4n}\right)+1\right)\\\\ &\times\left(\frac1{2n+1}-\frac{\pi}{2(2n+1)}\frac{\pi }{2(2n+2)}\int_0^1 x^{2n+2}\sin(\pi x/2)\,dx\right)\tag3 \end{align}$$

Using $\lim_{n\to \infty}\frac{\frac12\left(\cot\left(\frac{\pi}{4n}\right)+1\right)}{\frac{2n}{\pi}}=1$ and letting $n\to \infty$ and letting $n\to \infty$ in $(3)$ yields the intended limit

$$\lim_{n\to\infty}\left(\sum_{k=1}^n\sin\left(\frac{\pi k}{2n}\right)\int_0^1 x^{2n}\sin(\pi x/2)\,dx\right)=\frac1\pi$$

Answer 2

Method 1 (Laplace's method).

We use the asumptotics of $$\int^1_0 x^{2n} \sin\left( \frac{\pi x}{2}\right)\,dx=\int^1_0 e^{2n(\log(x))} \sin\left( \frac{\pi x}{2}\right)\,dx $$ Via Laplace's method one gets as $n\to\infty$: $$ \int^1_0 x^{2n} \sin\left( \frac{\pi x}{2}\right)\,dx \sim \frac{1}{2n } $$ The required limit is thus: \begin{align} \lim_{n\to\infty} \sum_{k=1}^n \frac{\pi k}{2n} \int^1_0 x^{2n} \sin\left( \frac{\pi x}{2}\right)\,dx = \lim_{n\to\infty} \frac{\pi}{2n} \cdot \frac{(n+1)n}{2} \cdot \frac{1}{2n} = \frac{\pi}{8} \end{align} where we have used the well known result $\sum_{k=1}^n k = \frac{n(n+1)}{2}$.

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Method 2 (Elementary).

So let me derive this more elementary. Set $t=1-x$ to get: $$\int^1_0 x^{2n}\sin\left( \frac{\pi x}{2}\right)\,dx = \int^1_0 (1-t)^{2n} \cos\left( \frac{\pi t}{2}\right)\,dt $$ Recall the inequality $1-\frac{x^2}{2} \leq \cos(x) \leq 1$. Hence: \begin{align} \int^1_0 (1-t)^{2n}\left(1- \frac{\pi^2}{8}t^2\right)\,dt \leq \int^1_0 (1-t)^{2n} \cos\left( \frac{\pi t}{2}\right)\,dt\leq \int^1_0 (1-t)^{2n}\,dt \end{align} In other words: \begin{align} \frac{1}{2n+1} - \frac{\pi^2}{8}\int^1_0 (1-t)^{2n}t^2\,dt \leq \int^1_0 (1-t)^{2n} \cos\left( \frac{\pi t}{2}\right)\,dt\leq \frac{1}{2n+1} \end{align} Moreover $$\int^1_0 (1-t)^{2n}t^2\,dt = \frac{1}{(2n+1)(2n+3)(n+1)} = O(n^{-3})$$ which can be easily gotten by integration by parts (or through Beta function). All in all we conclude: \begin{align} \frac{(n+1)\pi}{4} \frac{1}{(2n+1)}+O(n^{-2})\leq \sum_{k=1}^n \frac{\pi k}{2n} \int^1_0 x^{2n} \sin\left( \frac{\pi x}{2}\right)\,dx \leq \frac{(n+1)\pi}{4} \frac{1}{(2n+1)} \end{align} where we (again) have used the formula for $\sum_{k=1}^n k$. With the Squeeze Theorem we conclude the same as in method 1.

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Method 3 (DCT).

This method is inspired by Mark Viola's excellent answer. Integrate by parts once to get: $$\int_0^1 x^{2n}\sin\left(\frac{\pi x}{2}\right)\,dx=\frac1{2n+1}-\frac{\pi}{2(2n+1)}\int_0^1 x^{2n+1}\cos\left(\frac{\pi x}{2}\right)\,dx$$ So: $$\sum_{k=1}^n \frac{\pi k}{2n} \int^1_0 x^{2n} \sin\left( \frac{\pi x}{2}\right)\,dx = \frac{(n+1)\pi}{4} \frac{1}{(2n+1)}-\frac{(n+1)\pi}{4}\frac{\pi}{2(2n+1)}\int_0^1 x^{2n+1}\cos\left(\frac{\pi x}{2}\right)\,dx $$ Taking the limits as $n\to\infty$ together with DCT ( $|x^{2n+1}\cos\left(\frac{\pi x}{2}\right)|\leq 1$ and it converges a.e. to $0$) yields the same result.