# Proof that $a_{n}:=\sup\{|\frac{nx}{1+n^2x^2}|:x\in \mathbb{R}|\}$ $(n\in \mathbb{N})$ does not converge to $0$

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, $$\lim_{n\rightarrow\infty} \sup\big\{|\frac{nx}{1+n^2x^2}|:x\in \mathbb{R}|\big\}>0$$.

• 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 $a_n$. Oct 20, 2022 at 19:39
• 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. Oct 20, 2022 at 19:42
• @FShrike . It is incorrect, as $b$ depends on $x$ and $x$ depends on $n$. We have $b<1/x^2<1/n^2.$ Oct 21, 2022 at 2:12

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

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

• 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 $0$? Oct 20, 2022 at 19:59
• @Charles, Well that's what i did, i chose $x_n=\frac{1}{n}$. Oct 20, 2022 at 20:03
• I mean using my chain of inequalities. I suppose not, since my inequalities require $x>n$, and so $b_n$ is necessarily converging to $0$. Oct 20, 2022 at 20:05
• @Charles, See my edit 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 $$0$$.

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