= Solution
A <Cauchy sequence> is a sequence $(x_n)$ such that for every $\varepsilon>0$ there is $N$ for which
$$
m,n\geq N\quad\Longrightarrow\quad |x_m-x_n|<\varepsilon.
$$
The general principle of convergence, or <completeness of the real numbers>, states that a real sequence converges if and only if it is Cauchy.
For the final claim, let $m=\lfloor n/2\rfloor$. Since $(a_n)$ is decreasing and positive,
$$
0\leq (n-m)a_n
\leq\sum_{k=m+1}^na_k.
$$
Because the <series> $\sum a_k$ converges, its tails tend to zero. Also $n-m\geq n/2$, so
$$
0\leq na_n
\leq2\sum_{k=m+1}^na_k
\longrightarrow0.
$$
The squeeze theorem gives
$$
\boxed{na_n\to0}.
$$
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