uniform convergence of $f_n(x)=1/n^x$

Question

Consider the sequence of functions: \begin{align} f_n(x) = \frac{1}{n^x}. \end{align}

Does $(f_n)_n$ converge uniformaly on $[0,\infty)$. I proved via the Cauchy criterion that $(f_n)_n$ can not converge uniformaly on $(-\infty, 0)$, but unfortunately could not study the uniform convergence on $[0,\infty)$.

I appreciate any help. Thanks in advance.

Answer 1

I don't think it converges uniformly on $ [0, \infty) $. Because if it does then we have $$ \lim_{n \to \infty} \lim_{x \to 0^+} f_n(x) = \lim_{x \to 0^+} \lim_{n \to \infty} f_n(x). $$ But a direct computation shows that the left hand side is 1 while the right hand side is 0, hence a contradiction.

Answer 2

No, it doesn't even converge uniformly on $(0,\infty)$. Suppose by way of contradiction that it does, let $\epsilon>0$ be given, then there exists an $N \in \mathbb{N}$ such that for all $n \geq N$, $|\frac{1}{n^{x}}|<\epsilon$ for all $x \in (0,\infty)$. Take $x=-\frac{\ln(\epsilon)}{\ln(N)}$ and pick $n=N$, then $\frac{1}{n^{x}}=\epsilon$ (needs to be strictly less than $\epsilon$). Contradiction.