Does $f_n(x) = \sqrt[n]{1+x^n}$ converge uniformly on $[0,\infty)$?

Question

Does $f_n(x) = \sqrt[n]{1+x^n}$ converge uniformly on $[0,\infty)$?

First, let's investigate point-wise convergence:

• For $x>1$: $\lim_{n\to\infty} \sqrt[n]{1+x^n} = x$
• For $x\le 1$: $\lim_{n\to\infty} \sqrt[n]{1+x^n} = 1$

So there is a point-wise convergence only for $x\ge 1$.

Basically, the question asks if there's a uniform convergence at $[0,\infty)$ so the answer is no because there isn't even a point-wise convergence at $[0,1)$. Am I right?

But (assuming I'm right so far) let's assume the question asked about uniform convergence at $[1,\infty)$:

We should look for $$\sup_{x\in [1,\infty)} \left( \sqrt[n]{1+x^n} -x \right)$$

Let's find extremum points of $f_n(x)$:

$$f_n'(x) = (1+x^n)^{\frac{1}{n}-1}x^{n-1}$$

Here I got a little stuck, tough I think there's only minimum point at $x=0$ so $f_n(x)$ grows with $x$.

Could you help me finish up this proof? (And validate the first part.)

Answer 1

Partial answer.

Use $$(u-v)(u^{n-1}+u^{n-2}v+\cdots + uv^{n-2}+v^{n-1}) = u^n-v^n$$ With $u=\sqrt[n]{1+x^n}$ and $v=x$. This will give you a good estimate for $f_n(x)-x$ when $x\geq 1$. In particular, $f_n(x)-x \leq \frac{1}{n}$ for $x\geq 1$.

So the question reduces to whether there is a uniform convergence on $[0,1)$.

Answer 2

The sequence of functions $$ f_n(x) = (1+x^n)^{\frac{1}{n}} $$ uniformly converges to the function $g(x)=\max(1,x)$ over $\mathbb{R}^+$. If $x\in[0,1]$ we have: $$0\leq f_n(x)-1 \leq 2^{\frac{1}{n}}-1\leq \frac{1}{n} \tag{1}$$ through the inequality: $$ u>v>0 \Longrightarrow u^{\frac{1}{n}}-v^{\frac{1}{n}}\leq \frac{u-v}{n v^{\frac{n-1}{n}}}\tag{2}$$ that also gives, over $[1,+\infty)$, $$ 0\leq f_n(x)-x \leq \frac{1}{n x^{n-1}}. \tag{3}$$