# Show that this sequence of functions converges uniformly to $e^{-x}$

Define

$$f_n(x)= \begin{cases} \left(1-\frac xn\right)^n & 0\le x

I am able to show that $$f_n(x)\to e^{-x}$$ pointwise on $$[0,+\infty)$$, and I wonder if my justification for uniform convergence is valid:

By properties of logarithm, I transformed the sequence into

$$\left(1-\frac xn\right)^n=\exp\left\{n\log\left(1-\frac xn\right)\right\}$$

Employing the fact that for sufficiently large $$n$$

$$\left|n\log\left(1-\frac xn\right)+x\right|=\left|n\left[-\frac xn+\mathcal O\left(1\over n^2\right)\right]+x\right|=\mathcal O\left(\frac1n\right)$$

we have $$n\log\left(1-\frac xn\right)\to-x$$ uniformly, hence for each $$\delta>0$$ there exists an $$N$$ such that for all $$n>N$$:

$$|\log f_n(x)+x|<\delta$$

By the uniform continuity of the exponential function, for all $$\varepsilon>0,x>0,y>0$$, we have $$|e^{-x}-e^{-y}|<\varepsilon$$ whenever $$|x-y|<\delta$$. As a result let $$y$$ be $$-\log f_n(x)$$, so for all $$\varepsilon>0,x>0$$ there exists an $$N$$ such that for all $$n>N$$

$$|f_n(x)-e^{-x}|<\varepsilon$$

I wonder if this proof is correct or please point out my mistake. Thank you all!

• In this line here, $\left|n\log\left(1-\frac xn\right)+x\right|=\left|n\left[-\frac xn+\mathcal O\left(1\over n^2\right)\right]\right|=\mathcal O\left(\frac1n\right)$, I'm not sure why you wrote $O\left(\frac1n\right)$ at the end. – Connor Harris Nov 10 '20 at 3:52
• @ConnorHarris I edited the question, there should be an $+x$ in the absolute value sign. – TravorLZH Nov 10 '20 at 5:16

Your proof is not valid. Your estimates are valid only for fixed $$x$$ and there is no uniformity. In particular $$n \log (1-\frac x n)$$ does not tend to $$-x$$ uniformly: If $$|n log (1-\frac x n) +x| <\epsilon$$ for $$n \geq n_0$$ and for all $$x you get a contradiction by letting $$x \to n$$.

Hint for a valid proof: Choose $$T$$ such that $$x>T$$ implies $$e^{-x} <\epsilon$$. Note that $$0 \leq f_n(x)\leq e^{-x} <\epsilon$$ for all $$n$$ if $$x>T$$. So it is enough to consider $$\sup_{x \leq T} |f_n(x)-e^{-x}|$$ and here your approach can be used.

• Did you mean that all I need is to show $f_n(x)\to e^{-x}$ uniformly on $[0,T]$ – TravorLZH Nov 10 '20 at 7:59
• Your estimates are uniformly valid on $[0,T]$ so you should have no difficulty in completing the proof. @TravorLiu – Kavi Rama Murthy Nov 10 '20 at 8:01
• Okay, that helps a lot – TravorLZH Nov 10 '20 at 8:03
• Is it that choosing $T$ satisfying some quantity being $<\varepsilon$ the common way to determine uniform convergence of some sequence of functions on $[0,+\infty)$? – TravorLZH Nov 10 '20 at 8:35
• Not always. It works in this case. @TravorLiu – Kavi Rama Murthy Nov 10 '20 at 8:37

I think you need to look closer at what's going on in the Taylor series. You can clean up the argument that $$f_n(x):=n\log\left(1-\frac{x}{n}\right) \to -x$$ uniformly, but this only works if $$x\leq M$$ for some $$M>0$$. To see this:

By Taylor's theorem, we know $$f_n(x/n)=n\underbrace{f_n(0)}_{=0}+\underbrace{f_n'(0)}_{=-1}x+\frac{f_n''(c)}{2}\left(\frac{x^2}{n}\right)$$ for some $$c$$ between $$0$$ and $$x/n$$. Hence, \begin{align} |f_n(x/n)+x|\leq \left|\frac{f_n''(c)}{2}\right|\left(\frac{|x|^2}{n}\right)\leq \left|\frac{f_n''(c)}{2}\right|\left(\frac{M^2}{n}\right). \end{align} An easy calculation shows $$f_n'(x)=\frac{-n}{1-\frac{x}{n}},\, f_n''(x)=\frac{-1}{(1-\frac{x}{n})^2}$$. Now if $$x$$ is not demanded to be bounded above by some $$M>0$$ independent of $$n$$, then clearly $$f_n''(c)$$ is unbounded. However, with this demand, we have $$|f_n''(c)|\leq \frac{1}{(1-\frac{M}{n})^2}\leq C<\infty$$ From here, your argument about the exponential being uniformly continuous allows you to assert that $$e^{f_n(x)} \to e^{-x}$$ uniformly on $$[0,M]$$.

Finally, given $$\epsilon>0$$, pick $$M$$, such that $$e^{-x}<\epsilon$$ for all $$x>M$$. Then for sufficiently large $$n$$, it is obvious that we have uniform convergence.