<s>[INST] How do I show that the arc swept by $(1+\frac{i}{n})^n$ has length 1? I'm working on a tutorial for Euler's identity, and trying to show that the sum of the lengths of the arrows in this picture converges to 1 as $n \rightarrow \infty$

The length of the bottom arrow is $\frac{1}{n}$, and each arrow gets longer by a factor of
$$\left|1 + \frac{i}{n}\right| = \sqrt{1 + \frac{1}{n^2}}$$
so, the total length is the geometric sum
$$
\lim_{n \rightarrow \infty} \sum_{k=0}^{n-1} \frac{\left(\sqrt{1 + \frac{1}{n^2}}\right)^k}{n}
=
\lim_{n \rightarrow \infty} \frac{1 - \left(\sqrt{1 + \frac{1}{n^2}}\right)^n}{\left(1 - \sqrt{1 + \frac{1}{n^2}}\right)n}
$$
I've fiddled with this a lot and I can't seem to get it into a form where I can take the limit.
 [/INST] As $n\to \infty$ we have $ \frac{1}{ n^2}\to 0$
As you know as $x\to 0$ we have $\sqrt{1+x}\approx 1+\frac{x}{2}$ thus
as $n\to \infty$ we have $\sqrt{1 + \frac{1}{n^2}}\approx 1+\frac{1}{2n^2}$
$\lim_{n \to \infty} \frac{1 - \left(\sqrt{1 + \frac{1}{n^2}}\right)^n}{\left(1 - \sqrt{1 + \frac{1}{n^2}}\right)n}=\lim_{n \to \infty}\frac{1-\left(1+\frac{1}{2n^2}\right)^n}{n\left(1-1-\frac{1}{2n^2}\right)}=\lim_{n \to \infty}\frac{1-\left(1+n\frac{1}{2n^2}+\ldots\right)}{-\frac{1}{2n}}=\lim_{n \to \infty}\frac{-\frac{1}{2n}}{-\frac{1}{2n}}=1$
Terms in parenthesis $(\ldots)$ go to $0$ when $n\to\infty$ because they are $\dfrac{const}{n^k}$
Hope this helps
</s>