# Uniform convergence of a sequence $f_n(x)=n/(x+n)$

Show that the sequence $\{f_n\}$ of function where $f_n(x)=n/(x+n)$, is uniformly convergent in $[0,k]$ whatever $k$ may be, but not uniformly convergent in $[0,\infty)$.

The sequence is point wise convergent $\forall x\geq 0$, $$f(x)=1\hspace{1 cm} \forall x\geq0$$

How to proceed further?

• Start with calculating $f(x) - \frac{n}{n+x}$. Look at the result. Oct 19 '17 at 19:18
• Actually, $\lim\limits_{x\to +\infty} \frac{x}{x+n} = 1$. What is the maximal value that $\frac{x}{x+n}$ attains on $[0,k]$? Oct 19 '17 at 19:37
• @DanielFischer $k/(k+n)$ Oct 19 '17 at 19:45
• No, you're confusing the roles of $k$ and $n$. Maybe it would have been better to us $a$ instead of $k$. You have $$0 \leqslant 1 - \frac{n}{x+n} \leqslant \frac{k}{k+n}$$ on $[0,k]$. Given $\varepsilon > 0$, you want $\frac{k}{k+n} \leqslant \varepsilon$ for all large enough $n$. Oct 19 '17 at 20:07

The pointwise limit is $f(x)=1$.
In $[0,k]$ we have $$|f(x)-f_n(x)|=\left|1-\frac{n}{x+n}\right|=\frac{x}{x+n}$$
The map $$x\mapsto\frac{x}{x+n}$$ is an increasing function on $[0,k]$. So the maxima of this function (which occurs at $x=k$) is $\frac{k}{k+n}$.
So $$\lim_{n\to\infty}(\sup_{x\in[0,k]}|f(x)-f_n(x)|)= \lim_{n\to\infty}\frac{k}{k+n}=0$$ Hence $f_n\rightarrow f$ uniformly in $[0,k]$.