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In my course on analysis, we defined the exponential function in terms of power series, that is, as $ \exp(x) = \sum_{n = 0}^{\infty} \frac{ x^n}{n!}$.

I now need to show that $$\lim_{n \to \infty} \big( 1 + \frac{x}{n} \big)^n = \exp x $$ for all $x \in \mathbb{R}$. Also, I need to determine if this equality still holds for $x \in \mathbb{C}$.

Attempt: I want to show that the difference of the terms above equals zero. We have \begin{align*} \lim_{n \to \infty} \bigg( 1 + x/n \bigg)^n - \sum_{n = 0}^{\infty} \frac{ x^n}{n!} = \lim_{n \to \infty} \sum_{k=0}^{n} \binom{n}{k} \frac{ x^k}{n^k} - \sum_{n=0}^{\infty} \frac{ x^n}{n!} \end{align*} which equals \begin{align*} \lim_{ n \to \infty} \sum_{k=0}^n \frac{ n!}{(n-k)! k!} \frac{x^k}{n^k} - \lim_{ n \to \infty} \sum_{ k=0}^n \frac{ x^k}{k!} \end{align*} or \begin{align*} = \lim_{ n \to \infty} \bigg[ \sum_{k=0}^n \frac{ x^k}{k!} \bigg( \frac{ n!}{(n-k)! n^k} - 1 \bigg) \bigg] \end{align*} Now for this expression to equal zero, I want to show that $$ \lim_{ n \to \infty} \frac{ n!}{(n-k)! n^k} = 1. $$ And this is the part where I'm stuck. Also, I don't know if the above equality holds in $\mathbb{C}$, but I presume not. Help is appreciated!

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    $\begingroup$ You could use Stirling approximation $\endgroup$ – openspace Apr 16 '16 at 8:47
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    $\begingroup$ my prefered way : $f_n(x) = (1+x/n)^n$, $f_n'(x) = (1+x/n)^{n-1}$, when $n \to \infty$ : $\frac{1}{1+x/n} = \frac{f_n'(x)}{f_n(x)} \to 1$ uniformly on $|x| < M$ hence $\ln f_n(x) \to x$ uniformly, hence $f_n(x) \to e^x$ uniformly $\endgroup$ – reuns Apr 16 '16 at 8:57
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Note that $\frac{n!}{(n-k)!}$ is - after cancelling - just the product of $k$ integers in the range $n-k+1$ upt to $n$. Hence $$\frac{(n-k)^k}{n^k}\le \frac{n!}{(n-k)!n^k}\le \frac{n^k}{n^k}=1. $$ But clearly for fixed $k$ $$ \lim_{n\to\infty}\frac{(n-k)^k}{n^k}=\lim_{n\to\infty}\left(1-\frac kn\right)^k=1.$$

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  • $\begingroup$ Thanks. That was very clear. Do you have any hint on whether the above equality holds in $\mathbb{C}$? $\endgroup$ – Kamil Apr 16 '16 at 9:06
  • $\begingroup$ @Kamil If you think your proof does not work for $x\in \Bbb C$, you may not have made yourself clear enough why it works for $x\in\Bbb R$. $\endgroup$ – Hagen von Eitzen Apr 16 '16 at 11:29

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