If $(a_n)$ is a decreasing sequence of strictly positive numbers and if $\sum{a_n}$ is convergent, show that $\lim{na_n}=0$ [duplicate]

Suppose that $\lim\limits_{n\to\infty}na_n\ne0$. That is, there is an $\epsilon\gt0$ so that for any $N$, there is an $n\ge N$ so that $na_n\ge\epsilon$. Since $a_n$ is decreasing, $$ \sum_{k=n/2}^n a_n\ge\frac{n}{2}\frac{\epsilon}{n}=\frac{\epsilon}{2} $$ Since $na_n$ does not converge to $0$, we can find infinitely many such blocks from $n/2$ to $n$ that sum to at least $\epsilon/2$. Thus, we've shown the contrapositive: if $\lim\limits_{n\to\infty}na_n\ne0$, then $\sum\limits_{k=0}^\infty a_n$ diverges.

Hint Use Cauchy's criterion for the convergence of a series. Remember the sequence is decreasing and positive.