An algebra problem by Trishit Chandra

Algebra Level 4

$\large \lim_{n\to\infty} \dfrac{\pi}2 \sum_{j=1}^{2^n} \sin\left( \dfrac{j \pi}2 \right) = \, ?$

0 1 4 2

1 solution

Chew-Seong Cheong
Nov 13, 2019

For integer $n > 1$ , $2^n$ is a multiple of 4 and we have:

$\begin{aligned} \sum_{j=1}^{2^n} \sin \left(\frac j2\pi\right) & = 1 + 0 - 1 + 0 +1 + \cdots + 0 + 1 + 0 - 1 + 0 = 0 \end{aligned}$

Therefore, $\displaystyle \lim_{n \to \infty} \frac \pi 2 \sum_{j=1}^{2^n} \sin \left(\frac j2\pi\right) = \boxed 0$ .