Uniform convergence of $f_n(x) = n(x^{1/n} - 1)$ to $f(x) = \ln x$ on each interval $(1/m, m), m \in \mathbb{N}$

Question

The function $f_n : (0, \infty) \longrightarrow \mathbb{R}$ is defined by $$f_n(x) = n(x^{1/n}-1)$$ Show that {$f_n$} converges pointwise to $f(x) = \ln x$. Show that the convergence is uniform on each interval $(\frac{1}{m}, m), m \in \mathbb{N}$, but not on $(0, \infty)$.

I've done the point wise part. But can not do the uniform convergence. I've tried using the :
$\sup\left\{|f_n - f| : x \in (1/m, m)\right\}$ $\rightarrow 0$ as $n \rightarrow 0$
I saw that supremum will be at $x \rightarrow n$, so the expression will be evaluated to $g(n) = n(n^{1/n}-1)- \ln n$. If i'm not wrong now my goal would be, to show the sequence $g(n) \rightarrow 0$ as $n \rightarrow 0$.
It would be generous if somebody could help me out. Thanks in advance.

Answer 1

First of all you are using bad notations. You want to prove uniform convergnce of $(f_n)$ on $(\frac 1m, m)$ for any $m$ and the way you are using $n$ with two meanings is not correct.

Now you will see that what is needed is uniform convergence $n(x^{1/n}-1)-\ln x$ to $0$. differentiating this show that the function $n(x^{1/n}-1)-\ln x$ is decreasing on $(\frac 1m , 1)$ and increasing on $(1,m)$. So it is enough to prove that

a) $n(m^{1/n}-1) -\ln m \to 0$

b) $n(m^{-1/n}-1) +\ln m \to 0$

Both these follow from the fact $n(t^{1/n}-1)=n(e^{\frac 1n \ln t}-1)\sim n(\frac 1n \ln t +0(\frac 1 {n^{2}})) \to \ln t$ with $t=m$ and $t=\frac 1 m$.

Suppose the convergence is uniform on $(0,\infty)$. Then there exist $n_0$ such that $|n(x^{1/n}-1)-\ln x | <1$ for all $x >0$ for all $n \geq n_0$. Fix $n=n_0+1$ and take limit as $x \to \infty$. You will get a contradiction because the left side tends to $\infty$ as $x \to \infty$.

[$n(x^{1/n}-1)-\ln x=\ln x [ n\frac {x^{1/n}} {\ln x} -1]-n$. Apply L'Hopital's Rule to see that $\frac {x^{1/n}} {\ln x} \to \infty$].

Answer 2

You are right, that we need to investigate $\lim\limits_{n \to \infty}\sup\limits_X|n(x^{\frac{1}{n}}-1)-\ln x|$. It's easy to reach with derivative: $$\big( n(x^{\frac{1}{n}}-1)-\ln x\big)'=x^{\frac{1}{n}-1}-\frac{1}{x}$$ Now it's easy to see, that on each $(a,b)$, which not contain $0$, this function is bounded and limit gives zero. But in neighbourhood of $0$ and $\infty$ it's not bounded.

Answer 3

Take $m\in\Bbb N$ and $\varepsilon>0$; you want to prove that there is a $N\in\Bbb N$ such that$$(\forall n\in\Bbb N)\left(\forall x\in\left(\frac1m,m\right)\right):n\geqslant N\implies\left|n\left(\sqrt[n]x-1\right)-\log x\right|<\varepsilon.$$Note that if $h\in(0,\infty)$ and $x>0$, then\begin{align}\frac{x^h-1}h-\log x&=\frac{e^{h\log x}-h\log x-1}h\\&=\log(x)\frac{e^{h\log x}-h\log x-1}{h\log x}.\end{align}Since$$\lim_{h\to0}\frac{e^h-h-1}h=0,$$there is a $\delta>0$ such that$$|h|<\delta\implies\left|\frac{e^h-h-1}h\right|<\frac\varepsilon{\log m}.$$Now, take $N\in\Bbb N$ such that $N>\frac{\log m}\delta$. Then, if $n\geqslant N$ and if $x\in\left(\frac1m,m\right)$,\begin{align}\left|n\left(\sqrt[n]x-1\right)-\log x\right|&=\left|\frac{x^{1/n}-1}{1/n}-\log x\right|\\&=\log(x)\left|\frac{e^{\log(x)/n}-\log(x)/n-1}{\log(x)/n}\right|\\&<\varepsilon,\end{align}since$$\left|\frac{\log x}n\right|\leqslant\frac{\log m}N<\delta.$$