Proof of uniform convergence of $f_n(x) = \dfrac{x^n}{1+x^n}$ on every interval $[b, \infty[$ with $b>1$

Question

I can only proof this for $b>2$.

Then, you can find $N\in\mathbb{N}$ so that $\left|\dfrac{x^n}{1+x^n} -1\right| = \dfrac{1}{1+x^n} \leq \dfrac{1}{n}< \varepsilon$ for all $n\geq N$ and for all $x\in [b, \infty[$.

But what to do, when $1<b<2$?

Answer 1

hint

For $b>1$ and $x\ge b$

$$|\frac{x^n}{1+x^n}-1|=|\frac{1}{1+x^n}|$$

$$\le \frac{1}{b^n}$$

thus

$$\sup_{x\in [b,+\infty)}|f_n(x)-1|\le \frac{1}{b^n}$$

You can conclude.

Answer 2

Hint: $$\forall x \in [b,\infty[:1>\dfrac{x^n}{1+x^n}=1-\dfrac{1}{1+x^n}>1-\dfrac{1}{1+b^n}$$