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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 $n \rightarrow \infty$

enter image description here

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.

Will Orrick
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jedediah
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  • What is i, is it the imaginary unit ? – gary Jul 16 '17 at 01:05
  • Yeah, sorry, should have mentioned that – jedediah Jul 16 '17 at 01:09
  • The limit expression you have falls to the squeeze thm. Note that $1+n^{-2} > 1$. Thus, the sum you have is lower bounded by $1$, and upper bounded by $\frac{n (1 + n^{-2})^{n/2}}{n} = (1 + n^{-2})^{n/2}.$ Since $(1 + x)^{1/x} \le e,$ we have $(1 + n^{-2})^{n/2} \le e^{1/2n} \to 1$. – stochasticboy321 Jul 16 '17 at 04:51
  • Do I understand correctly that $z=\left(1+\frac{i}{n}\right)^n$ is the value at the tip of each arrow? I tried plotting that starting with $n=1$ and got nothing like that. – Cye Waldman Jul 16 '17 at 23:01
  • @CyeWaldman The value at the tip of each arrow is $\left(1+\frac{i}{n}\right)^k$, where $k$ ranges from $1$ (for the arrow just above the horizontal axis) to $n$ (for the last arrow in the counterclockwise direction). The goal is to compute the value of the last arrow, which is $\left(1+\frac{i}{n}\right)^n$. – Will Orrick Jul 16 '17 at 23:34
  • @CyeWaldman Sorry, the last sentence I wrote was wrong. The goal is to compute the sum of the lengths of the arrows. The complex multiplication $\left(1+\frac{i}{n}\right)^k$ can be built up by noticing that $z\left(1+\frac{i}{n}\right)$ is a complex number whose magnitude is $\left|v\right|\sqrt{1+\frac{1}{n^2}}$ and whose argument is that of $z$ plus $\arctan\frac{1}{n}$. The figure consists of a series of right triangles; the length of the long leg of one triangle is the length of the hypotenuse of the previous. (That's the $\left|z\right]$ mentioned above.) The short leg of the... – Will Orrick Jul 17 '17 at 00:09
  • ...first triangle has length $\frac{1}{n}$. In each subsequent triangle, the length of the short leg grows by a factor $\sqrt{1+\frac{1}{n^2}}$. – Will Orrick Jul 17 '17 at 00:11

4 Answers4

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HINT: This polygonal path approaches the arc of the unit circle from $(1,0)$ to $e^i = (\cos 1,\sin 1)$.

Ted Shifrin
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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 $\frac{1}{n}$.

Will Orrick
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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

Raffaele
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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.

$$\lim_{n\to\infty}s=\frac{2n^2}{n}(1+\frac{n}{2n^2}-1)=1$$

Cye Waldman
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