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Show that $ \left( 1 + \frac{x}{n}\right)^n$ is uniformly convergent on $S=[0,1]$.

Given $f_n(x)=\left( 1 + \frac{x}{n}\right)^n$ is a sequence of bounded function on $[0,1]$ and $f:S \rightarrow \mathbb R$ a bounded function , then $f_n(x)$ converges uniformly to $f$ iff

$$\lim\limits_{n \rightarrow \infty} ||f_n - f|| = \lim\limits_{n \rightarrow \infty}\left(\sup |f_n(x) - f(x)| \right)= 0$$

As $$f(x) = \lim\limits_{n \rightarrow \infty} \left( 1 + \frac{x}{n}\right)^n = e^x$$

We have

$$\begin{split} \lim\limits_{n \rightarrow \infty} \left\|\left( 1 + \frac{x}{n} \right)^n - e^x\right\| &= \lim\limits_{n \rightarrow \infty} \left|\sup_{x \in S}\left[\left( 1 + \frac{x}{n}\right)^n - e^x \right ]\right|\\ &= \lim\limits_{n \rightarrow \infty} \left|\left( 1 + \frac{1}{n} \right)^n - e^1 \right|\\ &= \lim\limits_{n \rightarrow \infty} |e-e| \\ &= 0 \end{split} $$

I am new to sequence. Is this appropriate to show convergence?

Going back to the definition, How can I show that:

  1. "$f_n(x)=\left( 1 + \frac{x}{n}\right)^n$ is a sequence of bounded function on $[0,1]$"?

  2. $f:S \rightarrow \mathbb R$ a bounded function?

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    $\begingroup$ Continuous functions are bounded on closed intervals. For any $n$, $f_n(x)$ expands to a polynomial, which is continuous anywhere. $\endgroup$ – IntegrateThis Aug 14 '17 at 0:15
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    $\begingroup$ The third line in your last calculation is wrong: shouldn't have a limit in front of $|e-e|$. $\endgroup$ – symplectomorphic Aug 14 '17 at 1:23
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    $\begingroup$ @Will Fisher: well, sure, as written it doesn't make a difference. But the point is that it in preserving the limit from the second to the third line it looks like the student is arguing that $|(1+1/n)^n-e|$ equals $|e-e|$. $\endgroup$ – symplectomorphic Aug 14 '17 at 1:27
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    $\begingroup$ @Will Fisher: no, you have misunderstood the OP. The OP is proving that $f_n\to f$ uniformly by invoking hypotheses, and the questions at the bottom are about how to show the hypotheses hold. $\endgroup$ – symplectomorphic Aug 14 '17 at 1:29
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    $\begingroup$ Have you shown that the sup obtains when $x=1$? It is true, but it seems to me to require proof (and you have the absolute value in the wrong place: as it stands - the absolute value of the sup of the difference-, the sup is 0 at $x=0$; you want the sup of the absolute value of the difference instead). $\endgroup$ – NickD Aug 14 '17 at 1:51
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If you don't want to verify the precise maximum you can bound as

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

since $e^{x/n} > 1 + x/n$ which implies $e^{x} > (1 + x/n)^n $ for $x \in (-n,\infty).$

By Bernoulli's inequality, $(1 - x^2/n^2)^n \geqslant 1 - x^2/n$ and

$$0 \leqslant e^x - \left(1 + \frac{x}{n} \right)^n \leqslant \frac{e^xx^2}{n} \leqslant \frac{e}{n},$$

which enables you to prove uniform convergence on $[0,1]$.

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Let's write explicitly the difference $$e^x - ( 1 + \frac{x}{n})^n = \sum_{ k \ge 0} \frac{x^k}{k!} - \sum_{k = 0}^n \binom{n}{k}\left(\frac{x}{n}\right )^k=\\ =\sum_{k\ge 0} \frac{[1-\prod_{l=1}^{k-1}(1- l/n)] x^k}{k!}$$. Since every coefficient $1- \prod_{l=1}^{k-1} (1-\frac{l}{n})$ is $\ge 0$ ( and $0$ for $k \ge n+1$ ) we conclude that

$$0 < e^x - (1+ \frac{x}{n})^n\le e^1 - (1 + \frac{1}{n})^n$$ for all $n \ge 0$ and $x\in [0,1]$.

Now, it's only necessary to check ( or use ) that $(1+\frac{1}{n})^n \to e$.

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As shown in inequality $(2)$ of this answer, $\left(1+\frac xn\right)^n$ is increasing in $n$. Thus, for $x\ge0$, $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\left(e^x-\left(1+\frac xn\right)^n\right) &=e^x-\left(1+\frac xn\right)^{n-1}\\ &\ge e^x-\left(1+\frac xn\right)^n\\[3pt] &\ge0 \end{align} $$ So on $[0,1]$, $$ 0\le e^x-\left(1+\frac xn\right)^n\le e-\left(1+\frac1n\right)^n $$

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The logarithm $\ln:[1,e]\to[0,1]$ is uniformly continuous on this interval, so it is a uniform homeomorphism, and it suffices to show that $\ln f_n(x)=n\ln (1+x/n)$ is uniformly convergent on $[0,1]$. But this follows immediately from Taylor's theorem: $$ n\ln(1+x/n)-x=O(x^2/n). $$

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