Prove or disprove: if $\sum\limits^{\infty} a_n$ is convergent, then $\lim\limits_{n \to \infty}na_n=0.$ The question comes from the following problem:
If$\lim\limits_{n \to \infty}(a_1+a_2+\cdots+a_n)=S,$ then$\lim\limits_{n \to \infty}\dfrac{a_1+2a_2+\cdots+na_n}{n}=0.$
I want to use Stolz theorem. Thus, I obtain $$\lim\limits_{n \to \infty}\dfrac{a_1+2a_2+\cdots+na_n}{n}=\lim\limits_{n \to \infty}(n+1)a_{n+1}.$$
If we can prove $ \lim\limits_{n \to \infty}na_n=0,$ the problem is solved. But I wonder whether it holds or not.
 A: Try the alternating harmonic series:
$$\sum_{k=1}^\infty \frac{(-1)^k}{k}.$$
If you assume $a_n$ is monotone, then it is true (and is a fun exercise to prove).
A: Set
$$a_n=\left\{ 
\begin{array}{lc}
\frac{1}{k^2} &\mbox{ if } n= k^2 \\
0 & \mbox{ otherwise }
\end{array}
\right.$$
OR
$$b_n=\left\{ 
\begin{array}{lc}
\frac{1}{n} &\mbox{ if } n= k^2 \\
\frac{1}{n^2} & \mbox{ otherwise }
\end{array}
\right.$$
A: By using the literature, as suggested by user @i707107, it is possible to find proofs that there are many successions of coefficients $\langle a_n \rangle_{n \in \mathbb{N}}$ such that
$$
\sum_{n=1}^{\infty} a_n=L<\infty\text{ does not imply }\lim_{n \to \infty} na_n=0\iff a_n=o(n)
$$
Precisely, the basic theorem on alternating series (see for example [1], pp. 156-157) states that every series of the form
$$
\sum_{n=1}^{\infty} a_n=\sum_{n=0}^\infty (-1)^nb_n,
$$
where $b_n\ge 0$ for all $n\in\mathbb{N}$ is a strictly monotonically decreasing sequence, converges, no matter how slow is the vanishing rate of the coefficients, and also 
$$
R_m=\left|\sum_{n=m+1}^{\infty} a_n\right|=\left|\sum_{n=m+1}^\infty (-1)^nb_n\right|<b_{m+1}=|a_{m+1}| \quad \forall m\in\mathbb{N}
$$
Therefore, any arbitrarily chosen strictly and slowly monotonically decreasing succession of positive numbers can be used to construct a counterexample to the main assertion of the question: for example the succession $a_n=(-1)^n \big[\log(\log(n+1))\big]^{-1}$ satisfies the requirements.
However, the sought for result, i.e.
$$
\lim\limits_{n \to \infty}(a_1+a_2+\cdots+a_n)=s \text{ implies }\lim\limits_{n \to \infty}\dfrac{a_1+2a_2+\cdots+na_n}{n}=0.
$$
is true: you can look at the proofs of Tauber's second theorem, for example the one offered by Jan Van de Lune [2], §1.2 p. 4, to see how to proceed.
[1] Emanuel Fisher (1983), "Intermediate Real Analysis", Springer Verlag
[2] Jan van de Lune, (1986), "An introduction to Tauberian theory: from Tauber to Wiener", CWI Syllabus, 12, Amsterdam: CWI, pp. iv+102, ISBN 90-6196-309-5, MR 0882005, Zbl 0636.40002
