Prove uniform convergence for $f_n(x) = \sqrt[n]{x^2+xn + 1}$ on $(0,1)$ and $(1, +\infty)$

Question

By what methods and means I can prove that function's UC on these intervals?

I know the answer, it uniformly converged on first interval and not uniformly on second to $f(x) = 1$, but I need to prove it somehow.

Answer 1

Using Bernoulli's inequality or binomial expansion, we have for $n > 1$

$$1 + x^2 + xn =f_n(x)^n = [1 + (f_n(x) - 1)]^n \\> 1 + n(f_n(x) - 1) + \frac{n(n-1)}{2}(f_n(x) - 1)^2 \\> 1 + \frac{n(n-1)}{2}(f_n(x) - 1)^2.$$

Hence,if $x \in (0,1)$

$$0 \leqslant |f_n(x) - 1| < \sqrt{\frac{2x(x+n)}{n(n-1)}} < \sqrt{\frac{2(1+n)}{n(n-1)}} = O(1/\sqrt{n}).$$

The RHS converges to $0$ as $n \to \infty$. Hence, for any $\epsilon > 0$ we have $|f_n(x) - 1| < \epsilon$ for all $x \in (0,1)$ and $n$ sufficiently large. Thus , convergence is uniform on $(0,1)$.

Non-uniform convergence on $(1,\infty)$ is relatively obvious. Try to find a sequence $x_n \in (1,\infty)$ such that $|f_n(x_n) - 1| \not\to 0.$