if $\lim_{n\to\infty}a_n=0$ then exists $p$ such that $\sum_{n=1}^\infty (a_n)^p$ converges?

Question

Is it true or false? To me it seems true, because if $\lim_{n\to\infty}a_n=0$ then $(a_n)^p$ approaches 0 much faster, and if we will take a $p$ that is large enough, it will approach even faster than $\frac{1}{n^2}$ (and $\sum_{n=1}^\infty \frac{1}{n^2}$ converges) but I can't seem to prove it.. any help will be appreciated !

Answer 1

If $a_n = \frac{1}{\log n}$ (for $n \geq 2$) then $\sum_{n=2}^\infty a_n^p$ diverges for all $p$ since $\frac{1}{\log^p n} = \Omega(\frac{1}{n})$.