Uniform convergence of $f_n (x) = \frac{1+x^n}{n+x^{2n}}$ on $\mathbb{R}$

Question

I found the sequence pointwise converges to $f(x) = 0$ if $x \in (-1, + \infty)$, now I have to say if it uniformly converges and in order to do it I think I have to find a sup of $f_n$ on $(-1,+\infty)$

Answer 1

Yes, it converges uniformly. The only zero of $f_n'$ is $\displaystyle\sqrt[n]{\sqrt{1+n}-1}$. Furthermore, $f'(x)<0$ when $x$ is greater than this number and $f'(x)>0$ if $x$ is smaller. Therefore, the maximum of $f_n$ is$$f_n\left(\sqrt[n]{\sqrt{1+n}-1}\right)=\frac1{2\left(\sqrt{1+n}-1\right)}.$$Since$$\lim_{n\in\mathbb N}\left(\frac1{2\left(\sqrt{1+n}-1\right)}\right)_{n\in\mathbb N}=0,$$your sequence converges uniformly to the null function.

Answer 2

Yes. the sequence $\{f_n\}_{n\geq 1}$ converges uniformly to $0$ in $(-1,+\infty)$: as $n\to \infty$, $$\sup_{x\in (-1,+\infty)}|f_n(x)|= \sup_{x\in [1,+\infty)}\frac{1+x}{n+x^{2}}=\frac{1+x_n}{n+x_n^{2}}\to 0$$ where $x_n=\sqrt{1+n}-1$ (consider the derivative of $\frac{1+x}{n+x^{2}}$).