opepayflarpws

2022-06-26

Limit with a summation and sine: how to calculate $\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\sum _{k=0}^{n-1}\mathrm{sin}\left(\frac{2\pi k}{n}\right)$ ?

laure6237ma

Expert

$\sum _{k=0}^{n-1}\mathrm{sin}\left(2\pi k/n\right)$ is the imaginary part of $\sum _{k=0}^{n-1}{e}^{i2\pi k/n}$ . We can evaluate the latter sum directly since it's a truncated power series:
$\sum _{k=0}^{n-1}{z}^{k}=\frac{{z}^{n}-1}{z-1}$
provided that $z\ne 1$, so
$\sum _{k=0}^{n-1}{e}^{i2\pi k/n}=\frac{{e}^{i2\pi }-1}{{e}^{i2\pi /n}-1}=\frac{1-1}{{e}^{i2\pi /n}-1}=0$
Hence also
$\sum _{k=0}^{n-1}\mathrm{sin}\left(2\pi k/n\right)=0$
for every positive integer n. So your limit is
$\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\sum _{k=0}^{n-1}\mathrm{sin}\left(2\pi k/n\right)=\underset{n\to \mathrm{\infty }}{lim}\left({n}^{2}\cdot 0\right)=\underset{n\to \mathrm{\infty }}{lim}0=0$

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