Convergent series multiplied by $n$ this one should be really simple...
Let $\mathbb a_n$ be a sequence of strictly positive real numbers such that $\sum_{n\in\mathbb N}a_n$ is finite, i.e., the limit $\lim_{k\to\infty}\sum_{n=1}^ka_n$ exists.
Then $a_n$ has to converge to zero, pretty quickly so. -- That much I do remember.
Does it follow that $\lim_{n\to\infty}a_n\cdot n=0$ has to hold?
Here is what I do (seem to) remember:
If the limit exists, then $a_n$ has to converge to zero, ``quicker'' than $1/n$,
thus(???), there has to exist some $\epsilon>0$ such that $a_n$ goes to zero as fast as  $(1/n)^{1+\epsilon}$.
--- But is this true?
If this is in fact true, then $\lim_{n\to\infty}a_n\cdot n = 0$ does hold.
 A: We have a counterexample by
$$a_n = \begin{cases}\frac{1}{n} &, n \text{ is a square}\\ \frac{1}{2^n} &, n\text{ is not a square}. \end{cases}$$
This sequence has $\limsup\limits_{n\to\infty} na_n = 1$, and
$$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac{1}{2^n} - \sum_{k=1}^\infty \frac{1}{2^{k^2}} + \sum_{k=1}^\infty \frac{1}{k^2} < \sum_{n=1}^\infty \frac{1}{2^n} + \sum_{k=1}^\infty \frac{1}{k^2} < \infty.$$
If we took a sparser sequence of special values, say $a_n = \frac{1}{\sqrt{n}}$ for $n$ a cube, we could also have $\limsup\limits_{n\to\infty} na_n = +\infty$.
However, if the sequence $a_n$ is monotonic, then $na_n \to 0$ follows.
By the Cauchy condensation test, if $a_n$ is monotonic,
$$\sum_{n=1}^\infty a_n < \infty \iff \sum_{k=0}^\infty 2^ka_{2^k} < \infty.$$
Since the series on the left converges by assumption, so does the right, and that implies that $\lim_{k\to\infty} 2^ka_{2^k} = 0$. But then for $2^k \leqslant n < 2^{k+1}$ we have
$$0 \leqslant n a_n \leqslant 2^{k+1}a_n \leqslant 2^{k+1}a_{2^k} = 2(2^ka_{2^k}),$$
and the right hand side converges to $0$. By squeezing, so does $na_n$.
