I am trying to prove that the sequence $a_{n}:=\sup\big\{|\frac{nx}{1+n^2x^2}|:x\in \mathbb{R}|\big\}$ $(n\in \mathbb{N})$ does not converge to $0$. Is the following correct?

Consider arbitrary $n\in \mathbb{N}$. Choose some $x>n$. Then $$|\frac{nx}{1+n^2x^2}|\geq |\frac{n^2}{1+n^2x^2}|\geq |\frac{n^2}{n^2+n^2x^2}|=|\frac{1}{1+x^2}|:=b>0$$

Thus, for any $n\in \mathbb{N}$, $\sup\big\{|\frac{nx}{1+n^2x^2}|:x\in \mathbb{R}|\big\}\geq b>0$ (as, otherwise, for any $n \in \mathbb{N}$ we could find an $x\in \mathbb{R}$ such that $|\frac{nx}{1+n^2x^2}|>\sup\big\{|\frac{nx}{1+n^2x^2}|:x\in \mathbb{R}|\big\}$—a contradiction. Thus, $\lim_{n\rightarrow\infty} \sup\big\{|\frac{nx}{1+n^2x^2}|:x\in \mathbb{R}|\big\}>0$.

  • $\begingroup$ It might be not that tedious to actually find the maxima of the functions $x\rightarrow|nx/(1+n^2x^2)$ and get a much detailed look at what is happening with $a_n$. $\endgroup$
    – Mittens
    Oct 20, 2022 at 19:39
  • 1
    $\begingroup$ The problem here is that your $b$ depends on the $x$ you chose which itself depends on $n$, meaning it's more of a $b_n$ that you don't really know about, the sequence $b_n$ could converge to $0$ itself (for example if the $x$ you use is $n+1$!). However, plugging a different $n$-related value as an $x$ does work. $\endgroup$
    – Bruno B
    Oct 20, 2022 at 19:42
  • $\begingroup$ @FShrike . It is incorrect, as $b$ depends on $x$ and $x$ depends on $n$. We have $b<1/x^2<1/n^2.$ $\endgroup$ Oct 21, 2022 at 2:12

2 Answers 2


As Bruno B suggested in the comments, it seems that $b$ depends on $n$. So the inequality that we would get is $a_n\geq b_n$, which is not sufficient as $b_n$ may tends to zero.

Also, since you chose $x_n$ such that $x_n>n$, we would get $b_n=\left|\frac{1}{1+x_n^2}\right|\leq \left|\frac{1}{1+n^2}\right|\xrightarrow{n\to\infty} 0$. So choosing $x$ such that $x>n$ would not work in this case.

Note that

$$ a_n=\sup\left\{\left|\frac{nx}{1+n^2x^2}\right|:x\in \mathbb{R}\right\}\geq \frac{n\cdot \frac{1}{n}}{1+n^2\cdot \frac{1}{n^2}}=\frac{1}{2} $$

Thus, because limit preserve weak inequality, if $a_n$ converges, it will uphold

$$ \lim_{n\to\infty} a_n\geq \frac{1}{2} $$

  • $\begingroup$ Thank you. But could my solution still work if I chose $x$, for each $n$, as to ensure that $b_{n}$ does not converge to $0$? $\endgroup$
    – Charles
    Oct 20, 2022 at 19:59
  • $\begingroup$ @Charles, Well that's what i did, i chose $x_n=\frac{1}{n}$. $\endgroup$
    – Udi Fogiel
    Oct 20, 2022 at 20:03
  • $\begingroup$ I mean using my chain of inequalities. I suppose not, since my inequalities require $x>n$, and so $b_n$ is necessarily converging to $0$. $\endgroup$
    – Charles
    Oct 20, 2022 at 20:05
  • $\begingroup$ @Charles, See my edit $\endgroup$
    – Udi Fogiel
    Oct 20, 2022 at 20:09

You can assume $x\ge0$. Let $$ f_n(x)=\frac{nx}{1+n^2x^2}. $$ Then $$ f_n'(x)=\frac{n(1-n^2x^2)}{(1+n^2x^2)^2}. $$ Letting $f_n'(x)=0$ gives $x=:x_n\equiv\frac1n$. Clearly $$ f_n''(x_n)=-\frac{n^2}{2}<0 $$ and hence $f_n(x)$ attains a local maximum at $x=x_n$. Since $f_n(0)=f_n(\infty)=0$, you can conclude that $f_n(x)$ attains the global maximum at $x=x_n$. Therefore $$ a_n=\sup\big\{|\frac{nx}{1+n^2x^2}|:x\in \mathbb{R}\big\}=f_n(x_n)=\frac12 $$ which implies $a_n$ do not converge to $0$.

  • 3
    $\begingroup$ (+1) That is what I had in mind in my comment to the OP. Not such a tedious computation after all. $\endgroup$
    – Mittens
    Oct 20, 2022 at 20:49

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