$\sum a_k$ convergent and $(a_k)$ decreasing monotonically $\Rightarrow n\cdot a_n\to 0$
Proof
Let $\varepsilon >0$. Given the series $\sum_{k=1}^{\infty}{a_k}$ converges, there is (using Cauchy's criterion) an Indexx $n_0$ such that for all $m>n>n_0$ we have:$$0\leq \sum \limits_{k=n+1}^{m}a_k\leq \frac{\varepsilon}{2}$$ Since $a_k$ decreases monotonically, we have $(m-n)a_m<\frac{\varepsilon}{2}$. $(*)$
Choosing $n_1=2n_0+1$, $m>n_1$ arbitrary and $n=\left \lfloor \frac{m}{2}\right \rfloor$, we have $n>n_0$ and thus:$$\frac{m}{2}a_m\leq (m-n)a_m<\frac{\varepsilon}{2}$$ Hence we have $m\cdot a_m<\varepsilon$ for all $m>n_1$ und since $\varepsilon$ was chosen arbitarily, we conclude $m\cdot a_m\to 0$.
What I don't understand is how we know $(*)$ - is this some kind of theorem?