Prove that the lim sup of this sequence is $\ge$ than $0$ Let $\sum a_n$ be a convergent series of arbitrary terms. If $p_n$ increases monotonely to $+\infty$ in such a way that $\sum \frac1{p_n}$ is divergent, then we have $$\limsup \frac{p_1a_1+p_2a_2+\ldots+p_na_n}{n} \ge 0$$
I only know that:
$\frac{p_1a_1+\ldots+p_na_n}{p_n} \to 0$ as $n\to +\infty$ by Kronecker's lemma, hence we need only to look at the case $\limsup \frac{p_n}{n} = +\infty$. I'm not sure if this is useful, but I think that we should apply summation by parts here, however I don't know how to use the fact that $\sum\frac1{p_n}$ is divergent. Can you give me a hint?
 A: Let's go for a contrapositive. We shall keep the assumption that $(p_n)$ is monotonically increasing (not necessarily strictly) and eventually positive. And, since we want a contrapositive, we shall assume that
$$\limsup_{n \to \infty} \frac{p_1a_1 + p_2a_2 + \ldots + p_na_n}{n} < 0\,.$$
Thus there is an $\varepsilon > 0$ and an $n_0$ such that for all $n > n_0$ we have
$$S_n := \sum_{k = 1}^{n} a_kp_k \leqslant -\varepsilon n\,.$$
We can assume that $n_0$ is so large that $p_n > 0$ for $n > n_0$. Now let $n > n_0$. We then have
\begin{align}
\sum_{k = n_0+1}^{n} a_k
&= \sum_{k = n_0+1}^{n} \frac{S_k - S_{k-1}}{p_k} \\
&= -\frac{S_{n_0}}{p_{n_0+1}} + \sum_{k = n_0+1}^{n} S_k\biggl(\frac{1}{p_k} - \frac{1}{p_{k+1}}\biggr) + \frac{S_n}{p_{n+1}} \\
&\leqslant -\frac{S_{n_0}}{p_{n_0+1}} -\varepsilon \Biggl[\sum_{k = n_0+1}^{n} k\biggl(\frac{1}{p_k} - \frac{1}{p_{k+1}}\biggr) + \frac{n}{p_{n+1}}\Biggr] \\
&= -\frac{S_{n_0}}{p_{n_0+1}} - \varepsilon\Biggl[\frac{n_0}{p_{n_0+1}} + \sum_{k = n_0+1}^{n} \frac{k - (k-1)}{p_k}\Biggr] \\
&= -\frac{S_{n_0}}{p_{n_0+1}} - \varepsilon\frac{n_0}{p_{n_0+1}} - \varepsilon \sum_{k = n_0+1}^{n} \frac{1}{p_k}\,.
\end{align}
From this inequality we deduce

*

*if $\sum a_n$ converges, then $\sum_{n > n_0} \frac{1}{p_n}$ converges too (and if we additionally assume that $p_n \neq 0$ for all $n$ we can drop the $n > n_0$ condition).

*if $\sum_{n > n_0} \frac{1}{p_n} = +\infty$, then $\sum a_n = -\infty$.

Thus we have proved two contrapositives, for dropping different parts of the hypothesis, and consequently we proved that if $\sum a_n$ converges, $(p_n)$ is monotonically increasing, never $0$, and $\sum \frac{1}{p_n} = +\infty$, then
$$\limsup_{n \to \infty} \frac{p_1a_1 + p_2a_2 + \ldots + p_na_n}{n} \geqslant 0\,.$$
