# Limit of $f_n(x) = n\log(\frac{nx+1}{nx-1})$ and its uniform covergence

Let $f_n(x) = n\log(\frac{nx+1}{nx-1})$. What is the pointwise limit of $f_n$ for $x \in[1,\infty]$? Is it uniformly convergent?

$$\lim_{n\longrightarrow \infty}n\log(\frac{nx+1}{nx-1})=\lim_{n\longrightarrow \infty}\frac{\log\left(\frac{x+\frac{1}{n}}{x-\frac{1}{n}}\right)}{\frac{1}{n}}\stackrel{H}{=}\lim_{n\longrightarrow \infty}\frac{\frac{-2x}{n^2}\frac{1}{(x-\frac{1}{n})(x+\frac{1}{n})}}{\frac{-1}{n^2}}=\lim_{n\longrightarrow \infty}\frac{2x}{(x-\frac{1}{n})(x+\frac{1}{n})}=\frac{2}{x}$$ Is that correct? And how to check uniform covergence? I would really like not to search for $\sup$ of $|n\log(\frac{nx+1}{nx-1}) - \frac{2}{x}|$...

• Could some explain to me how the log is gotten rid of in the third step? – Padraic Aug 13 '14 at 9:25
• Using de l'Hospital rule I calculate the derivative of $\log(\frac{nx+1}{nx-1})$ and $\frac{1}{n}$ and the fraction after "H" equality is what I get (technically I should first check if this limit exists, but fortunately it does in this case). – Jules Aug 13 '14 at 9:34

Looks right to me. The supremum check is pretty easy here. Note that,

\begin{align} g(x) &:= \frac 2x - n\log(\frac{nx+1}{nx-1}) \\ &= \frac 2x - n\log(nx+1) + n\log(nx-1) \\ \implies g'(x) &= -\frac 2{x^2} - \frac {n^2}{nx+1} + \frac {n^2}{nx-1} \\ &= -\frac 2{x^2} + \frac {2n^2}{(nx+1)(nx-1)} \\ &= -\frac 2{x^2} + \frac {2n^2}{n^2x^2-1} \\ &= -\frac 2{x^2} + \frac {2}{x^2-\frac 1{n^2}} \\ &\geq 0 \end{align}

So $g(x)$ is monotonic and approaches it's supremum at an endpoint. It's pretty clear that $\lim_{x\to\infty}g(x)=0$, so $\sup_{x\in[1,\infty)}|g(x)| = |g(1)| = |2-n\log(\frac{n+1}{n-1})| \underset{n\to\infty}{\longrightarrow} 0$

• But we don't know that $g(x)>0$. – agha Jun 15 '14 at 21:53
• @agha, that's why I said endpoint, otherwise it would get its supremum only at positive infinity. – davin Jun 15 '14 at 21:56

Your pointwise limit is correct (you could have checked that with Wolfram Alpha or such). I haven't checked it in full detail, the derivatives in the L'Hopital's rule step look like they might have a small error (one that goes away as $n \to \infty$, however).

To justify uniformity, you can compute the limit in a different way entirely. If we bring the $n$ inside the logarithm and do long division, then we get

$$\log \left ( \left ( \frac{nx+1}{nx-1} \right )^n \right ) = \log \left ( \left ( 1 + \frac{2}{nx-1} \right )^n \right ) = \log \left ( \left ( 1 + \frac{1}{n} \frac{2}{x-1/n} \right )^n \right )$$

You may recognize the way I've written the inside as converging to $\exp(2/x)$. So if you know how to show that $\left ( 1 + x/n \right )^n$ converges uniformly (on compact sets) to $\exp(x)$, then you can argue that the logarithm is uniformly continuous on $[1,\infty)$ to finish the problem.