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.

  • 2
    $\begingroup$ If $0<x<1$, then $x^{2n} \to 0$ so the numerator will diverge to $-\infty$ and the denominator will converge to $1$. Hence, the $f_n(x) \to -\infty$. Should it be $\ln(1+x^{2n})$? $\endgroup$
    – AlanD
    Nov 12, 2020 at 19:54
  • $\begingroup$ I think it is not $ln(1+x ^{2n})$, it would have been easier. I don't agree with $f_n (x) \to - \infty$. If you consider a fixed value of x, you have $f_n (x) = \frac {2n log(x)} {(1+x)^{2n}} $, which tends to zero, because in pointwise convergence I calculate the limit in $n$, and is $x$ is rather a parameter. I hope I'm not wrong. $\endgroup$ Nov 12, 2020 at 20:17
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    $\begingroup$ You must have written $f_n(x)$ incorrectly in the question above. Perhaps edit it. Your denominator is $(1+x^{2n})$. $\endgroup$
    – AlanD
    Nov 12, 2020 at 20:19
  • $\begingroup$ Yes, it was a typo. Sorry. The post is now edited. $\endgroup$ Nov 12, 2020 at 20:33

1 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 $n \log n \underset{n \to \infty} \longrightarrow +\infty$.

  • $\begingroup$ Can I say that $\sup_{x \in (0,1]}\left|\frac{2n\log x}{(1+x)^{2n} }\right| = +\infty $ only by using the fact that $lim_ {x \to 0} |f_n(x) |= +\infty$, or do I need something else? $\endgroup$ Nov 13, 2020 at 7:42
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    $\begingroup$ @AdrianoBanchieri: Yes that is correct since $(1+x)^{2n}$ is bounded on that interval with $n$ fixed. If this is bothering you I can give another argument. $\endgroup$
    – RRL
    Nov 13, 2020 at 16:16

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