# Evaluating $\lim_{n\rightarrow\infty}\left(\frac{\pi}{2^n}\left(\sum_{j=1}^{2^n}\sin\left(\frac{j\pi}{2^n}\right)\right)\right)$

Evaluate $$\lim_{n\rightarrow\infty}\left(\frac{\pi}{2^n}\left(\sum_{j=1}^{2^n}\sin\left(\frac{j\pi}{2^n}\right)\right)\right)$$

I separated the expressions and also evaluated the summation. My answer is coming $$0$$. Can anybody review$$?$$ The software wolframalpha is showing nothing.

Any help is greatly appreicated.

EDIT

Let $$S=\sum_{j=1}^{2^n}\sin\left(\frac{j\pi}{2^n}\right)$$ Considering $$\theta=\displaystyle\frac{\pi}{2^n}$$ and using the formula I got the value of $$S$$ $$S=\frac{\displaystyle\sin\left(\frac{2^n+1}{2}\cdot\displaystyle\frac{\pi}{2^n}\right)}{\displaystyle\sin\left(\frac{\pi}{2^n\cdot2}\right)}\cdot\sin\left(\frac{\pi\cdot2^n}{2\cdot2^n}\right)$$

Now, re-writing the limit as $$\pi\lim_{n\rightarrow\infty}S\cdot\lim_{n\rightarrow\infty}\frac{1}{2^n}$$ Now, the value of the last limit is $$0$$ so the final answer is $$0$$.

• Does it make a difference if you replace $2^n$ with $n$? Or is it just easier to deal with? Oct 5, 2022 at 17:13
• @courageousmartingale i think it won't make a difference but even in that case my answer is coming $0$. Since there is a $\frac{1}{\infty}$ term the answer will always be $0$ in my opinion. Oct 5, 2022 at 17:15
• Do you know about Riemann sums? Oct 5, 2022 at 17:23
• @StefanLafon oh no no...im just an amateur...but if you have a solution using that...I'll appreciate that Oct 5, 2022 at 17:26
• "Now, the value of the last limit is 0 so the final answer is 0." Only if the first limit exists and is finite. "Since there is a 1∞ term the answer will always be 0 in my opinion." What about $1 = \lim_{n\to \infty} 1 =\lim_{n\to \infty}\frac nn = \lim_{n\to \infty} n\cdot \frac 1n = \lim_{n\to \infty} n\cdot \lim_{n\to \infty} \frac 1n = \lim_{n\to\infty} n\cdot 0 = 0$. Do you see what (2 things) are wrong with that? Oct 5, 2022 at 19:39

You have wrongly assumed that $$\lim_{n\to\infty} S$$ is finite. Hence, you cannot use limit laws to write it like in your last line. Instead, \begin{align*} \lim_{n\rightarrow\infty}\left(\frac{\pi}{2^n}\left(\sum_{j=1}^{2^n}\sin\left(\frac{j\pi}{2^n}\right)\right)\right) &= \pi \left(\lim_{n\to\infty} \sin\left(\frac{2^n+1}{2}\cdot\displaystyle\frac{\pi}{2^n}\right)\right)\left(\lim_{n\to\infty}\frac{1}{2^n\sin\left(\frac{\pi}{2^{n+1}}\right)}\right). \end{align*} For the first term, you can take the limit inside because sine is a continuous function to get $$\sin(\pi/2)=1$$.
For the second term, I assume you already know $$\lim_{x\to0} \sin(x)/x=1$$. It should be easy to show that the second term gives $$2/\pi$$ using that.
As one of the comments suggested, you can use integration to solve this problem. For a continuous function $$f$$ on the interval $$[a, b]$$, we have \begin{align*} \int_a^b f(x)dx &= \lim_{n\to\infty} \frac{b-a}{n} \sum_{j=1}^n f\left(a+j\frac{b-a}{n}\right). \end{align*} So, your problem becomes $$\int_0^{\pi} \sin(x)dx = \cos(0)-\cos(\pi) = 2$$.