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I have shown that $f_n\left(x\right)=n\left(\sqrt[n]{x}-1\right)$ converges pointwise to $f\left(x\right)=\ln x$ on $\left(0,\infty\right)$, and I am now trying to show that it converges uniformly on $\left[e^{-A},e^A\right]$.

Recall the definition of uniform convergence:

Let $f_n$ be a sequence of functions defined on a set $A\subseteq\mathbb{R}$. Then, $\left(f_n\right)$ converges uniformly on $A$ to a limit function $f$ defined on $A$ if, for every $\epsilon>0$, there exists an $N\in\mathbb{N}$ such that $\left|f_n\left(x\right)-f\left(x\right)\right|<\epsilon$ whenever $n\geq N$ and $x\in A$.

One thing I noticed I could use is that if $x\in\left[e^{-A},e^A\right]$ and $y=\ln x$, then $y\in\left[-A,A\right]$. However, after having tried that out, I have run of ideas as to how to begin to tackle this problem altogether. Could you guys give me a few pointers? Thanks in advance!

Edit 1: I have tried using the triangle inequality: $$\left|n\left(\sqrt[n]{x}-1\right)-\ln x\right|\leq\left|n\left(\sqrt[n]{x}-1\right)\right|+\left|\ln x\right|.$$And since $x\in\left[e^{-A},e^A\right]$: $$\left|n\left(\sqrt[n]{x}-1\right)\right|+\left|\ln x\right|\leq\left|n\left(\sqrt[n]{x}-1\right)\right|+\left|A\right|=n\left(\sqrt[n]{x}-1\right)+A.$$ Does this seem right? If so, all that I am left to do is show that for any $\epsilon>0$, there exists an $N\in\mathbb{N}$ such that the last expression above is less than $\epsilon$ whenever $n\geq N$ and $x\in \left[e^{-A},e^A\right]$. But I am stuck. What do you guys think?

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up vote 2 down vote accepted

We use the fact that for a real number $t$: $e^t=\sum\limits_{k=0}^{+\infty}\frac{t^k}{k!}$. We have \begin{align*} |f_n(x)-\ln x|&=\left|n\exp\left(\frac{\ln x}n\right)-n-\ln x\right|\\ &=\left|n\sum_{k\geq 1}\frac{(\ln x)^k}{k!n^k}-\ln x\right|\\ &\leq n\sum_{k\geq 2}\frac{|\ln x|^k}{k!n^k}\\ &=n\sum_{j\geq 0}\frac{|\ln x|^{j+2}}{(j+2)!n^{j+2}}\\ &\leq \frac{|\ln x|^2}n\exp|\ln x| \end{align*} and if $x\in [e^{-A},e^A]$, we get $|f_n(x)-\ln x|\leq \frac{A^2e^A}n$, which gives uniform convergence.

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Could you clarify to me how $\sqrt[n]{x}=\exp\left(\frac{\ln x}{n}\right)$? – EsX_Raptor Mar 12 '12 at 14:35
It follows directly from the exponent laws. – azarel Mar 12 '12 at 14:41
To fill that out a bit if you haven't seen it before, notice that $e^{\log(x)} = x$ by definition and $\frac{1}{n} \log(x) = \log( x^{\frac{1}{n}})$ by elementary properties of logs. Put those together to get the result. – Chris Janjigian Mar 12 '12 at 14:58
I agree. My apologies. I immediately saw it when I remembered the exponent laws. I have another question: In the second step, I know that we start the summation at $1$ and not $0$ because the summation term at $0$ cancels out with the $n$. Similarly, in the third step, we can start the summation at $2$ and not $1$ because the summation term at $1$ cancels out with the $\ln x$. In the fourth step, I know that we change the index of the summation to revert back to the definition. However, I do not understand how we reached the fifth step. Did we use partial fraction decomposition? – EsX_Raptor Mar 12 '12 at 15:20
Never mind. I wrote out the terms of the sums and saw why we did that. Thank you very much, guys! – EsX_Raptor Mar 12 '12 at 16:01

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