Limit of complex sum

Question

$$ \lim_{n \to \infty} \sum_{k=1}^n \left| e^{2\pi i k /n} - e^{2\pi i (k-1)/n} \right| $$ Please help me with this one. I can't figure it out how to break the modulus as it's coming very large.

Answer 1

You can pull out $e^{2\pi ik /n}$ (which has modulus $1$) and simplify the sum as $n|1-e^{-2\pi i/n}|$. Use the fact that $e^{z}=1+z+o(z)$ as $x \to 0$ we see that the limit is $2\pi$.

More details: $e^{2\pi ik /n}-e^{2\pi i(k-1) /n}=e^{2\pi ik /n} [1-e^{-2\pi i/n}]$ so $|e^{2\pi ik /n}-e^{2\pi i(k-1) /n}|=|e^{2\pi ik /n}|| 1-e^{-2\pi i/n}|=| 1-e^{-2\pi i/n}|$. Now you see that the terms of this sum do not depend on $k$ at all! There are $n$ terms so you get $n | 1-e^{-2\pi i/n}|$. So what you have to find is limit of $n | [1-e^{-2\pi i/n}]|$. We can use the Taylor expansion of $e^{z}$ for this.

Answer 2

For a fixed $n$, the $e^{2\pi i k/n}$, $1 \leq k \leq n$, are the $n$ angles of a regular $n$-agon enscribed on the unit circle $z = 1$ in the complex plane. So intuitevely you can see that each of the term in the sum are equal because each term is equal to the lenght of a side of the polygon. This is indeed true :

$$ |e^{2\pi i k/n} - e^{2\pi i (k-1)/n}| = |e^{2\pi i (k-1)/n}||e^{2\pi i /n} - 1| =|e^{2\pi i /n} - 1|. $$

Now

$$ \begin{aligned} \sum_{k=1}^n |e^{2\pi i k/n} - e^{2\pi i (k-1)/n}| &= n |e^{2\pi i /n} - 1| = n \sqrt{ (\cos(2\pi /n) - 1)^2 + \sin^2(2\pi /n)} \\ &= n\sqrt{\cos^2(2\pi /n) - 2 \cos(2\pi /n) + 1 + \sin^2(2\pi /n) } \\ &= n\sqrt{2 (1-\cos(2\pi /n)} \\ &= 2n \sin(\pi /n) = 2\pi \frac{\sin(\pi /n)}{\pi n} \rightarrow 2\pi \end{aligned} $$

thanks to the formula $\cos(2x) = \cos^2(x) - \sin^2 (x) = 1 - 2\sin^2(x)$ and to the fact that $\frac{\sin x}{x} \rightarrow 1$ as $x$ tends to zero.

And this limit is what we should have expected: as $n$ goes to infinity the polygon has more and more sides so it approximates better and better the circle $z=1$ of circonference $2\pi$