Evaluate point convergence and uniform convergence of given funcion sequence

Question

I'm given a function sequence as such:

$$ f_n(x) = \frac{x^n}{n^n}$$ over $$x \in [0,1]$$

I have to find what is the point convergence and uniform convergence of this sequence. What is the proper method to evaluate this? Thanks in advance

Answer 1

Hint: For each $n\in\mathbb N$ and each $x\in[0,1]$, $\left\lvert\frac{x^n}{n^n}\right\rvert\leqslant\frac1{n^n}$.

Answer 2

Uniform convergence is in the supnorm, so $f_n(x) \rightarrow f(x)$ if $$ \sup_{x \in [0,1]} |f_n(x)-f(x)| \rightarrow 0 $$ Since $n^n$ is growing much faster than $x^n$, we can guess that $f_n(x) \rightarrow 0$. For uniform convergence, (notice that $x^n/n$ is increasing on $[0,1]$, so the largest distance between $x^n/n^n$ is attained at $x=1$) $$ \sup_{x \in [0,1]}|f_n(x)-0| = |1/n^n| $$ and $1/n^n$ \rightarrow 0, so $f_n \rightarrow_u f$.

Pointwise convergence is that, for all $x \in [0,1]$, $$ f_n(x) \rightarrow f(x). $$ So $f_n(x) \rightarrow 0$ for all $x \in [0,1]$, since $x^n/n^n \le 1/n^n$ for all $x \in [0,1]$, and $1/n^n \rightarrow 0$, so $f_n \rightarrow_p f$.