# Showing uniform convergence of $\frac{nx}{nx+1}$

Show that $$f_n(x) = \frac{nx}{nx+1}$$

converges uniformly to a function $$f$$ on domain $$[a, \infty)$$ where $$a>0$$.

Solution Verification Requested

Suppose $$f_n: [a, \infty) \to \mathbb{R}$$ is given by $$f_n(x) = \frac{nx}{nx+1}$$ .First notice that

$$f(x) = \lim_{n \to \infty}f_n(x) = \lim_{n \to \infty} \frac{nx}{nx+1} =1$$

Second, note that

$$|f_n(x) - f(x)| = \Big| \frac{nx}{nx+1} - 1 \Big| = \Big| \frac{nx}{nx+1} - \frac{nx+1}{nx+1} \Big| = \Big| \frac{-1}{nx+1} \Big| = \frac{1}{nx+1}$$

Now, let $$\varepsilon > 0$$. By the Archmidean Property, we can find $$N \in \mathbb{N}$$ so that

$$\frac{1}{Nx+1} < \varepsilon \hspace{1cm} x \in [a, \infty)$$

Suppose $$n \geq N$$. Since $$a \leq x$$, then $$\frac{1}{x} \leq \frac{1}{a}$$. Then we have

$$\frac{1}{nx+1} \leq \frac{1}{na+1}$$

Altogether, we have found

$$\frac{1}{nx+1} = |f_n(x)-f(x)| < \varepsilon$$

whenever $$n \geq N$$ and $$x \in [a, \infty)$$. So $$f_n \to f$$ uniformly.

As a final check, we check the endpoint $$a$$ and see that

$$|f_n(a+\frac{1}{n})-f(a+\frac{1}{n})| = \frac{1}{n(a+\frac{1}{n})+1} = \frac{1}{na+2} < \varepsilon \hspace{0.3cm} \mathrm{whenever} \hspace{0.3cm} n \geq N$$

• @ThomasAndrews My bad, I meant to add that in at the beginning but forgot. It's fixed now Commented Jun 23, 2022 at 19:13

I think that's perfect. Take a look to this form, maybe you'll find it easier, it uses the supremum criteiron $$\sup_{x \in [a,\infty)} \lvert f_n(x)-f(x) \rvert=\sup_{x \in [a,\infty)} \left\lvert \frac{1}{1+nx} \right\rvert=\frac{1}{1+n a}$$ and the latter tends to $$0$$ as $$n \to \infty$$.
Your appeal to the Archimedian property is dubious, because you are want it true for all $$x.$$ The reason you can do it is that $$\frac{1}{nx+1}$$ has a maximal value at $$x=a.$$ So just pick $$N$$ so that: $$\frac{1}{Na+1}<\epsilon.$$
I'd simplify it further, and pick $$N>\frac{1}{a\epsilon}.$$ Then for $$n\geq N, x\geq a,$$ $$nx\geq Na>0,$$ and thus $$\frac{1}{nx+1}<\frac{1}{nx}\leq\frac{1}{Na}<\epsilon.$$