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I need to show the following: $$\lim_{k\rightarrow \infty} \left(1 + \frac{z}{k}\right)^k=e^z$$

For all complex numbers z. I don't know how to start this. Should I use l'Hopitals rule somehow?

David K
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Dylan
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  • Perhaps you can recall what you do with the real number case – Vim Mar 12 '15 at 01:50
  • I've never seen this in the real number case. I've only seen a limit for e, not $e^z$ – Dylan Mar 12 '15 at 01:51
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    How are you defining $e^z$? – Milo Brandt Mar 12 '15 at 01:57
  • As eulers number e raised to some complex power? Im sorry I dont know what else you mean. – Dylan Mar 12 '15 at 02:00
  • This isn't really a strange result, although it is interesting when one sees it for the first time. – David K Mar 12 '15 at 02:14
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    @Dylan: there are many ways to define $e^z$, all equivalent. One of them is exactly what you are expected to prove. Another is to say $z=a+bi,\ e^z=e^a(\cos b + i\sin b)$. Another is to use the Taylor series. All are equivalent in that they give the same function. You pick one and have to prove the rest. People are asking where to start. – Ross Millikan Mar 12 '15 at 02:57
  • See this answer http://math.stackexchange.com/a/1668179/72031 This assumes that you define $e^{x+iy}$ as $e^{x}(\cos y + i\sin y)$. – Paramanand Singh Mar 25 '16 at 07:26
  • BTW, contrary to what you believe, raising a number to some complex power is not trivial. It requires reasonable amount of calculus machinery to define $a^{b}$ when $a,b$ are arbitrary real or complex numbers. – Paramanand Singh Mar 25 '16 at 07:31

3 Answers3

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Presumably you know that $\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)^n = e$. Try setting $n = \frac{k}{z}$ where $z$ is a constant and see where that gets you.

rogerl
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the following hints may suggest one line of approach. i will not spoil your fun by attempting to write a complete solution

if for positive integers $k \le m$ we set $$ f_{m,k}(z) = \binom{m}{k} \frac{z^k}{m^k} $$ (note that $\frac{\binom{m}{k}}{m^k} \lt \frac1{k!}$)

then for each $k$, we have $$ \lim_{m \to \infty} f_{m,k}(z) = \frac{z^k}{k!} $$ with the convergence being uniform on any disk $|z| \le R$

now define for all $m \gt n$: $$ F_{m,n}(z) = \sum_{k=0}^n f_{m,k}(z) $$ so that for each $n$ we have $$ F_n(z)=\lim_{m \to \infty} F_{m,n}(z) = \sum_{k=0}^n \frac{z^k}{k!} $$ with the convergence again uniform on compact disks.

the binomial theorem gives: $$ \left(1+\frac{z}{m} \right)^m = \sum_{k=0}^m \binom{m}{k} \frac{z^k}{m^k} $$ so for $m \gt n$ we may write $$ |(1+\frac{z}{m})^m - F_{m,n}(z)| \le \sum_{k=n+1}^m |f_{m,k}(z)| \lt \sum_{k=n+1}^m \frac{|z|^k}{k!} $$

David Holden
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The binomial expansion of $\left(1+\frac{z}{k}\right)^k$ is

$$\begin{align} \left(1+\frac{z}{k}\right)^k&=\sum_{j=0}^k{\binom{k}{j} \left(\frac{z}{k}\right)^j}\\\\ &=\sum_{j=0}^k \left( \frac{k!}{(k-j)!\,k^j}\right) \left(\frac{z^j}{j!}\right) \end{align}$$

The term $ \frac{k!}{(k-j)!\,k^j}$ can be expanded as

$$\begin{align} \frac{k!}{(k-j)!k^j}&=\frac{k(k-1)\dotsb (k-j+1)}{k^j}\\\\ &=\left(\frac{k}{k}\right) \left(\frac{k-1}{k}\right)\left(\frac{k-2}{k}\right) \dotsb \left(\frac{k-(j-1)}{k}\right)\\\\ &=\left(1-\frac1k\right)\left(1-\frac2k\right) \dotsb \left(1-\frac{j-1}{k}\right) \end{align}$$

and approaches $1$ as $k \rightarrow \infty$. Thus,

$$\begin{align} \lim_{k \to \infty} \left(1+\frac{z}{k}\right)^k&=\lim_{k \to \infty}\sum_{j=0}^k \left( \frac{k!}{(k-j)!\,k^j}\right) \left(\frac{z^j}{j!}\right)\\\\ &=\lim_{k \to \infty}\sum_{j=0}^\infty \xi_{[0,k]}(j)\left(\frac{k!}{(k-j)!\,k^j}\right) \left(\frac{z^j}{j!}\right)\\\\ & = \sum_{j=0}^\infty \frac{z^j}{j!} \end{align}$$

which is the series representation for $e^z$!

Note that $\xi_{[0,k]}(j)$ is the indicator function and is $1$ when $j\in[0,k]$ and $0$ otherwise. Also note that the last equality is justified since the series $\sum_{j=0}^\infty \xi_{[0,k]}(j)\left(\frac{k!}{(k-j)!\,k^j}\right) \left(\frac{z^j}{j!}\right)$ converges uniformly.

Mark Viola
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