Show that $(1+\frac{x}{n})^n \rightarrow e^x$ uniformly on any bounded interval of the real line.

I am trying to argue from the definition of uniform convergence for a sequence of real-valued functions, but am struggling quite a lot. My efforts so far have concentrated on trying to find a sequences, ${a_n}$ which tends to zero, such that

$$|(1+\frac{x}{n})^n -e^x |\leq a_n$$ for all $n$. But I have been unsuccessful thus far. All help is greatly appreciated.

  • 2
    $\begingroup$ The sequence is question is increasing and its limit is continuous. $\endgroup$ Nov 30 '16 at 1:44
  • $\begingroup$ Could you elaborate on your comment? I'm probably missing something simple. $\endgroup$ Nov 30 '16 at 1:47
  • $\begingroup$ There is a theorem that under the conditions in my comment the convergence is uniform on compact sets. $\endgroup$ Nov 30 '16 at 1:49
  • $\begingroup$ ...and said theorem goes by the name of Dini's theorem. $\endgroup$
    – Glitch
    Nov 30 '16 at 1:52
  • 2
    $\begingroup$ The sequence is not increasing on all of $\mathbb R$ so let's be careful. $\endgroup$
    – zhw.
    Nov 30 '16 at 2:07

Herein, we present an approach that for any given $\epsilon>0$, produces a number $N$, which depends on $\epsilon$ and not $x$, such that $\displaystyle \left|e^x-\left(1+\frac xn\right)^n\right|<\epsilon$ whenever $n>N$.

To do this we will use the inequalities, which I established in THIS ANSWER using only the limit definition of the exponential function and Bernoulli's Inequality. The inequalities used in the ensuing analysis are

$$\bbox[5px,border:2px solid #C0A000]{e^x\le 1+x} \tag 1$$

for $x<-1$ and

$$\bbox[5px,border:2px solid #C0A000]{\log(x)\ge \frac{x-1}{x}} \tag 2$$

for $x>0$.

To this end, we proceed.

We assume that $x\in [a,b]$ and that $\epsilon>0$ is given. Furthermore, we will choose $n$ such that $n>-x$ for all $x\in[a,b]$.

Using $(1)$ and $(2)$ we can write

$$\begin{align} \left|e^x-\left(1+\frac xn\right)^n\right|&=\left|e^x-e^{n\log\left(1+\frac xn\right)}\right|\\\\ &\le \left|e^x-e^{\frac{x}{1+x/n}}\right|\\\\ &=e^x\,\left|1-e^{-x^2/(x+n)}\right|\\\\ &\le e^x\,\left|\frac{x^2}{x+n}\right|\\\\ &\le e^{b}\frac{|\max^2(a,b)|}{n+a}\\\\ &<\epsilon \end{align}$$

whenever $ \displaystyle n>\frac{e^b\,\max^2(a,b)}{\epsilon}-a$. We take $\displaystyle N(\epsilon)=1+\left\lfloor \frac{e^b\,\max^2(a,b)}{\epsilon}-a \right\rfloor$ and we are done!

  • 1
    $\begingroup$ That's a clean approach +1 $\endgroup$
    – RRL
    Nov 30 '16 at 16:48
  • $\begingroup$ Your solution is fantastic, thank you so much. However, I don't completely follow you from the second to the third line (in which you pull out the $e^x$ form the absolute value. Should there be an equals sign there? I may have may be missing something but I can't seem to put that together. $\endgroup$ Dec 1 '16 at 0:55
  • $\begingroup$ You're welcome! And thank you! Recall that $e^x-e^y=e^x(1-e^{y-x})$. Now let $y=x/(1+x/n)$. $\endgroup$
    – Mark Viola
    Dec 1 '16 at 2:59
  • $\begingroup$ Of course, my mistake. Simple algebra and exhaustion don't mix well. Thanks again $\endgroup$ Dec 1 '16 at 4:31
  • 1
    $\begingroup$ @strants Thanks. I've edited. $\endgroup$
    – Mark Viola
    Dec 2 '16 at 5:01

To find your sequence $(a_n)$ where $|(1 + x/n)^n - e^x| \leqslant a_n \to 0$ -- proving uniform convergence on any bounded interval -- use the inequality $\ln(1+y) \leqslant y$.

We have for $0 \leqslant y < 1$,

$$1+y \leqslant e^y = \sum_{k=0}^{\infty} \frac{y^k}{k!} \leqslant \sum_{k=0}^{\infty} y^k = \frac1{1-y},$$

Take $y = x/n$. It follows that for $n$ sufficiently large

$$1 + \frac{x}{n} \leqslant e^{x/n} \leqslant \left(1 - \frac{x}{n}\right)^{-1},$$


$$\left(1 + \frac{x}{n}\right)^n \leqslant e^x \leqslant \left(1 - \frac{x}{n}\right)^{-n}.$$

The second inequality implies that

$$e^{-x} \geqslant \left(1 - \frac{x}{n}\right)^{n}.$$

Using Bernoulli's inequality $(1 - x^2/n^2)^n \geqslant 1 - x^2/n.$


$$0 \leqslant e^{x} - \left(1+ \frac{x}{n}\right)^n = e^{x}\left[1 - e^{-x}\left(1+ \frac{x}{n}\right)^{n}\right]\\ \leqslant e^{x}\left[1 - \left(1+ \frac{x}{n}\right)^{n}\left(1- \frac{x}{n}\right)^{n}\right]\\= e^{x}\left[1 - \left(1- \frac{x^2}{n^2}\right)^{n}\right]\leqslant e^{x}\frac{x^2}{n}.$$

Therefore, for all $x \in [0,K]$, we have as $n \to \infty$

$$0 \leqslant \left|e^{x} - \left(1+ \frac{x}{n}\right)^n\right| \leqslant e^K\frac{K^2}{n} \rightarrow 0.$$

An almost identical argument for $y \geqslant 0$ shows that

$$0 \leqslant e^{-y} - \left(1- \frac{y}{n}\right)^n \leqslant e^{-y}\frac{y^2}{n}.$$

Thus if $-y = x \in [-L,0]$ we have

$$0 \leqslant \left|e^{x} - \left(1+ \frac{x}{n}\right)^n\right| = \left|e^{-y} - \left(1- \frac{y}{n}\right)^n\right| \leqslant \frac{L^2}{n} \rightarrow 0.$$

proving uniform convergence on any bounded interval.

  • $\begingroup$ Your last assertion is incorrect: $f_n(x) = \left(1+\frac{x}{n}\right)^n$ is unbounded on any semi-infinite interval $(-\infty,K]$, so it cannot converge uniformly. (Specifically, your argument requires taking $n$ large relative to $|x|$ in line 2, which cannot be done on unbounded intervals). Otherwise, though, this is a solid argument, +1 $\endgroup$
    – user88319
    Dec 2 '16 at 4:26
  • $\begingroup$ @Strants: Thanks. You're correct. I'll take that out. $\endgroup$
    – RRL
    Dec 2 '16 at 5:00

If you're not familiar with Dini's theorem (I wasn't), work from the power series of $\exp$. Observe that $$(1+x/n)^n = \sum_{k=0}^n {n \choose k} \frac{x^k}{n^k}$$ and that $${n \choose k} \frac{1}{n^k} = \frac{n!}{n^k(n-k)!k!} \to \frac{1}{k!}$$

This means that each coefficient in the expansion of $g_n(x) = \exp(x) - \left(1+\frac{x}{n}\right)^n$ goes to $0$ as $n$ goes to infinity. Also, all of the coefficients are positive and less than $\frac{1}{k!}$.

Now, consider the interval $[-R,R]$. Choose $k> 0$ such that $\sum_{s=k+1}^\infty \frac{R^s}{s!} < \frac{\epsilon}{2}$ (why can you do this?) For $n$ sufficiently large, we can make each of the first $k+1$ coefficients smaller than $\frac{\epsilon}{2(k+1)(1+R^k)}$, so for any $x \in [-R,R]$, $$|g_n(x)| \le \sum_{s=0}^k \frac{\epsilon R^s}{2(k+1)(1+R^k)} + \sum_{s=k+1}^\infty \frac{R^s}{s!} \le \frac{\epsilon}{2} + \frac{\epsilon}{2}.$$

This same reasoning extends (with minor changes) to any bounded interval $[a,b]$ of $\mathbb{R}$ (and even half-open or open intervals).

  • $\begingroup$ This works. I found another approach that is straightforward, doesn't require splitting into cases, and leads directly to a number "$N(\epsilon)$." $\endgroup$
    – Mark Viola
    Nov 30 '16 at 16:38

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