Uniform Vs Pointwise convergence - tricky problem

Question

Consider $$ f_n (x) = \frac{log(x^{2n})} {(1 +x) ^{2n} } $$ for $x \in (0,1]$. What can I say about the convergence of this sequence of functions?

My attempt:

The sequence converges pointwise to the constant function 0, because $$ lim_{n \to \infty} f_n (x) =0$$ for $x \in (0,1]$. Now I don't know if the sequence converges uniformly. I suspect it doesn't. However, I get stuck trying to bound the supremum of $\vert f_n \vert$. I think it would be easier in a compact interval, e.g. $[a,1], 0\lt a \lt 1$. There, I would say $sup \vert f_n (x) \vert = \frac{2n\vert log (a) \vert} {(1+a)^{2n}} $, which tends to 0. Am I mistaken? Many thanks.

Answer 1

To confirm, for all $x \in (0,1]$, we have the pointwise convergence

$$\lim_{n \to \infty}\frac{\log x^{2n}}{(1+x)^{2n}} = \lim_{n \to \infty}\frac{2n\log x}{(1+x)^{2n} } =0$$

since $(1 +x)^n > nx $ and $\displaystyle\frac{n}{(1+x)^{2n}} < \frac{n}{n^2x^2} = \frac{1}{nx^2} \underset{n \to \infty} \longrightarrow 0$.

The sequence does not converge uniformly on $(0,1]$ since

$$\sup_{x \in (0,1]}\left|\frac{2n\log x}{(1+x)^{2n} }\right| = +\infty,$$

and, thus, $\lim_{n\to \infty} \sup_{x \in (0,1]} |f_n(x)| \neq 0$.

---

Alternatively, we can argue since $\frac{1}{n} \in (0,1]$ it follows that

$$\sup_{x \in (0,1]}\left|\frac{2n\log x}{(1+x)^{2n} }\right| = \sup_{x \in (0,1]}\frac{-2n\log x}{(1+x)^{2n} } \geqslant \frac{-2n\log \frac{1}{n}}{(1+\frac{1}{n})^{2n} } = \frac{2n\log n}{(1+\frac{1}{n})^{2n}} \geqslant \frac{2n\log n}{e^2},$$

and the limit of the LHS is $+\infty$ since $n \log n \underset{n \to \infty} \longrightarrow +\infty$.