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.


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 $0$.

  • $\begingroup$ Does it make a difference if you replace $2^n$ with $n$? Or is it just easier to deal with? $\endgroup$ Oct 5, 2022 at 17:13
  • $\begingroup$ @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. $\endgroup$
    – Vanessa
    Oct 5, 2022 at 17:15
  • $\begingroup$ Do you know about Riemann sums? $\endgroup$ Oct 5, 2022 at 17:23
  • $\begingroup$ @StefanLafon oh no no...im just an amateur...but if you have a solution using that...I'll appreciate that $\endgroup$
    – Vanessa
    Oct 5, 2022 at 17:26
  • $\begingroup$ "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? $\endgroup$
    – fleablood
    Oct 5, 2022 at 19:39

1 Answer 1


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.

Thus, the final answer is 2.

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 $\int_0^{\pi} \sin(x)dx = \cos(0)-\cos(\pi) = 2$.


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