Limit of $\lim\limits_{n\to\infty}\sum \limits_{k=n}^{2n}\sin(\frac{\pi}{k})$

Question

I'm asked to calculate the following limit: $$\lim\limits_{n\to\infty}\sum \limits_{k=n}^{2n}\sin(\frac{\pi}{k})$$ Of course, I can simply rewrite this to $$\lim\limits_{n\to\infty}\sum \limits_{k=0}^{n}\sin(\frac{\pi}{k+n})$$ Within the exercise, I was given the hint to interpret this as a Riemann integral.

Well unfortunately, I didn't get it. I've tried to use complex identity of $\sin(x)$:

$$\lim\limits_{n\to\infty}\sum \limits_{k=0}^{n}\frac{e^{\frac{i\pi}{k+n}}-e^{-\frac{i\pi}{k+n}}}{2i}$$

and Taylor series for $x_{0}=0$ $$\lim\limits_{n\to\infty}\sum \limits_{k=0}^{n} \lim\limits_{m\to\infty}\sum \limits_{l=0}^{m}(-1)^{l} \frac{(\frac{\pi}{k+n})^{2l+1}}{(2l+1)!}$$

But none of them let me recognize a Riemann sum.

If someone could give me a further hint, I would be really grateful!

Answer 1

Write $\sin x$ as a Taylor series instead:

$$\lim_{n\to\infty} \sum_{m=0}^\infty \frac{(-1)^m\pi^{2m+1}}{(2m+1)!} \sum_{k=0}^n \frac{1}{(k+n)^{2m+1}}$$

For the innermost summation, pull out the $n$:

$$\lim_{n\to\infty} \sum_{k=0}^n \frac{1}{\left(\frac{k}{n}+1\right)^{2m+1}}\cdot\frac{1}{n^{2m+1}}$$

For $m=0$ this is a Riemann sum. For $m > 0$, this is a Riemann sum times $n^{-2m}$. Provided the integrals converge (which they do, you will have to show this) these other limits go to $0$. Can you take it from here?

Answer 2

In a right neighbourhood of the origin we have $\sin(x)=x+O(x^3)$, so $$ \sum_{k=n}^{2n}\sin\frac{\pi}{k} = \pi\sum_{k=n}^{2n}\frac{1}{k}+O\left(\sum_{k=n}^{2n}\frac{1}{k^3}\right)\tag{1} $$ where $$ \sum_{k=n}^{2n}\frac{1}{k^3}\leq \int_{n-1}^{+\infty}\frac{dx}{x^3} = O\left(\frac{1}{n^2}\right) \tag{2}$$ $$ H_n=\sum_{k=1}^{n}\frac{1}{k} = \log(n)+\gamma+\frac{1}{2n}+O\left(\frac{1}{n^2}\right)\tag{3} $$ give us $$\sum_{k=n}^{2n}\sin\frac{\pi}{k} = \color{red}{\pi \log 2}-\frac{\pi}{4n}+O\left(\frac{1}{n^2}\right).\tag{4} $$