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.
 A: HINT: This polygonal path approaches the arc of the unit circle from $(1,0)$ to $e^i = (\cos 1,\sin 1)$.
A: You can also write your limit as
$$
\lim_{n \rightarrow \infty} \frac{1-\exp\left(\frac{n}{2}\log\left(1+\frac{1}{n^2}\right)\right)}{\left(1 - \sqrt{1 + \frac{1}{n^2}}\right)n}=\lim_{n\rightarrow\infty}n\left(1+\sqrt{1+\frac{1}{n^2}}\right)\left(\exp\left(\frac{n}{2}\log\left(1+\frac{1}{n^2}\right)\right)-1\right)
$$
Then Taylor expand the second factor in parentheses in powers of $\frac{1}{n}$.
A: 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
A: I just thought to chime in here with a slightly different point of view. For large $n$ we can take $z=\left(1+\frac{i}{n}\right)^k$ to be a continuous function, say
$$z=\left(1+\frac{i}{n}\right)^t$$
Now, the arc length in the complex plane is given by
$$s=\int |\dot z|~dt$$
First, let's simplify the algebra by letting $a=\left(1+\frac{i}{n}\right)$, then we can determine
$$
z=a^t\\
\dot z=\ln(a) ~a^t\\
|\dot z|=|\ln(a)|\cdot |a|^t,\quad |a|=\sqrt{1+\frac{1}{n^2}}
$$
Now we can construct the integral as follows
$$
\begin{align}
s
&=\int_0^n |\dot z|~dt\\
&=|\ln(a)|~\int_0^n |a|^t~dt\\
&=\frac{|\ln(a)|}{\ln (|a|)}|a|^t\biggr|_0^n\\
&=\frac{|\ln(a)|}{\ln (|a|)}(|a|^n-1)\\
\end{align}
$$
Now, noting that
$$\ln(1+x)=x-\frac{x^2}{2}+\cdots\\
(1+x)^n=1+nx+\cdots
$$
we can determine that for large $n$
$$
|\ln(a)|\approx \frac{1}{n}\\
\ln(|a|)\approx \frac{1}{2n^2}\\
|a^n|\approx 1+\frac{n}{2n^2}
$$
And finally.
$$\lim_{n\to\infty}s=\frac{2n^2}{n}(1+\frac{n}{2n^2}-1)=1$$
