Euler's Theorem Limit

Question

Does anyone know why $$\lim_{n\to ∞}\left(1+\frac {i\pi} n\right)^n=-1$$?

From my own intuition, subbing in $n=∞$ means that the $\frac x n \to 0$ and therefore the limit will be 1 as opposed to -1. I'm stumped at this part. Any help is greatly appreciated!

Answer 1

I think I've figured it out, please tell me if there is any problems with it (just a technicality, I'm only doing it for rational $n$ as I think that the continuity of $(1+\frac {i\pi}n)^n$ implies the limit for any real $n$):

$$e^{i\pi}=\lim_{n \to ∞} \left (1+ \frac {i\pi} n \right) $$ Using rewriting $1+\frac {i\pi} n$ as: $$\sqrt {1 + \frac {\pi^2} {n^2}}^n \left(\cos(\arctan(\frac {\pi} n) +i\sin(\arctan(\frac {\pi} n) \right)^n$$

Then using De Moivre's formula on this rewritten number (which can be proven by induction for integer $n$, and then for rational $n$, without Euler's formula), $$\left(1+ \frac {i\pi} n\right)^n=\sqrt {1 + \frac {\pi^2} {n^2}}^n \left(\cos(\arctan(\frac {\pi} n) +i\sin(\arctan(\frac {\pi} n) \right)^n$$ $$=\sqrt {1 + \frac {\pi^2} {n^2}}^n \left(\cos(n\arctan(\frac {\pi} n) +i\sin(n\arctan(\frac {\pi} n) \right)$$

As $n \to ∞$, the moduli raised to the $n$-th power $\to 1$ and the limit for $n \arctan ({\pi} n)$ will turn into $\pi$.

Plug $\pi$ into the $\cos$ and $\sin$ will give the answer of $-1$.

$$\therefore e^{i\pi}= \lim_{n \to ∞} \left (1+\frac {i\pi} n \right) =-1$$

Reference: See Post #12

Answer 2

This is $e^{i\pi}$, which is, by Euler's formula, $-1$.

Answer 3

To prove $\lim_{n\to \infty} (1+\frac{x}{n})^n=e^x$, use $(1+\frac{x}{n})^n=\sum_{k=0}^n\binom{n}{k} \frac{x^k}{n^k}$.
For each $k$, $\binom{n}{k} \frac{x^k}{n^k}=\frac{n(n-1)...(n-k+1)}{n^k} \frac{x^k}{k!}$

$\frac{n(n-1)...(n-k+1)}{n^k}$ is increasing with $n$ and $\to 1$ as $n\to \infty$
This leads to the series for $e^x$ which is $\sum_{k=0}^\infty \frac {x^k}{k!}$.

Answer 4

One can prove that $$\lim_{n\to \infty} \left(1+\frac{ix}{n}\right)^n=\cos x +i\sin x$$ for all $x\in \mathbb {R} $. Your desired result then follows by putting $x=\pi$.

Easiest route to this formula is the following lemma of MSE user Thomas Andrews:

Lemma: If $\{a_n\} $ is a sequence of complex numbers such that $n(a_n-1)\to 0$ as $n\to\infty$ then $a_n^n\to 1$ as $n\to\infty $.

For the current problem use $$a_n=\dfrac{1+\dfrac{ix}{n}}{\cos\dfrac{x}{n}+i\sin\dfrac {x} {n}} $$ and show that $n(a_n-1)\to 0$. Then $a_n^n\to 1$ and our job is done.