Determine whether $f_n(x)=\frac{x^n}{\sqrt{3n}}$ for $x \in [0,1]$ is uniformly convergent

Question

This is a follow up to my below question:

Pointwise convergence: State $f(x) = \lim f_n(x)$

I'm trying to determine whether $f_n(x)=\frac{x^n}{\sqrt{3n}}$ for $x \in [0,1]$ is uniformly convergent.

I know if it is then it must satisfy $\left|f_n(x)-f(x)\right|\lt \epsilon$

Since I know $f(x)=0$ I know it must be

$\left|f_n(x)-f(x)\right|\lt \epsilon = \left|\frac{x^n}{\sqrt{3n}}-0\right|\lt \epsilon =\left|\frac{x^n}{\sqrt{3n}}\right|\lt \epsilon$

After doing some algebra I get:

$\frac{1}{\epsilon}\lt \frac{\sqrt[n]{3n}}{x}$

Where do I go from here and have I made any mistakes? Does this means that $f$ is uniformly convergent?

Answer 1

Please I suggest you have a look at the following path. Since $x \mapsto x^n$ is increasing over $[0,1]$, then as $n \to \infty$, $$ \sup_{x \in [0,1]}|f_n(x)-0|=\sup_{x \in [0,1]}\left|\frac{x^n}{\sqrt{3n}}\right|=\frac{1}{\sqrt{3n}} \to 0, $$ then the sequence of functions $\left\{f_n\right\}_{n\ge1}$ is uniformly convergent over $[0,1]$.