Converse for Kronecker's lemma on infinite series Background
One formulation of Kronecker's lemma is the following.  Suppose $\{p_n\}$ is an increasing sequence of non-negative real numbers with $p_n \to \infty$ as $n \to \infty$. If $\sum a_n$ converges, then
\begin{equation}
\tag{1}\label{1}
\lim_{n \to \infty} \frac{1}{p_n} \sum_{k=1}^n p_ka_k = 0.
\end{equation}
Standard proofs of Kronecker's lemma proceed by summation by parts to get
\begin{equation}
\tag{2}\label{2}
\frac{1}{p_n} \sum_{k=1}^n p_ka_k = s_n - \frac{1}{p_n} \sum_{k=1}^{n-1}(p_{k+1}-p_k)s_k,
\end{equation}
where $s_n = \sum_1^n a_k$ denotes the sequence of partial sums.  For every $n$, 
$\frac{1}{p_n} \sum_{k=1}^{n-1}(p_{k+1}-p_k) = 1$ since the series telescopes.  The weighted averages on the right in \eqref{2} then converge to the same limit as the partial sums $\{s_n\}$ by Cesaro convergence, giving \eqref{1}.
This result is usually proved in probability textbooks due to its applications for strong laws of large numbers.
Question
I am trying to prove that if the limit \eqref{1} holds for every increasing non-negative sequence $\{p_n\}$ with $p_n \to \infty$ as $n \to \infty$, then $\sum a_n$ converges.
Note that this question is related, but not quite the same. Assuming that $\sum a_n$ diverges, I would want to show the existence of $\{p_n\}$ such that \eqref{1} fails.
Attempts
I tried considering the cases whether or not $\{s_n\}$ is bounded separately, and constructing $\{p_n\}$ accordingly to show that the weighted average of the $s_k$ does not get arbitrarily close to $s_n$, but couldn't quite close the arguments.
I would appreciate any solutions or hints.
 A: My attempt was on the right track, but was missing a key idea or two.  Here is the outline.
Suppose $\sum a_n$ diverges. Denote the sequence in (2) by $\{x_n\}$. We construct the sequence $\{p_n\}$ such that some subsequence $\{x_{n_j}\}$ does not converge to $0$.
Note that $\limsup s_n \ne \liminf s_n$.  We consider separately the three cases where $\limsup s_n = \infty$, $\liminf s_n = -\infty$, and $\limsup s_n$ and $\liminf s_n$ are both finite. (Note that these cases are not mutually exclusive, but are exhaustive.)
In each of these cases, pick out a subsequence $\{s_{n_j}\}$ converging to $\limsup s_n$ or $\liminf s_n$ (as appropriate), and define $\{p_n\}$ by $p_0 := 0$ and
$$p_{n+1} := \begin{cases}p_n+1 &\text{if } n=n_j \text{ for some }j,\\p_n &\text{otherwise}.\end{cases}$$
Then $\{p_n\}$ is nonnegative and increases to $\infty$. Moreover, the last term in (2) becomes the equally-weighted average of the subsequence $\{s_{n_j}\}$,
$$\frac{1}{p_n}\sum_{k=1}^n(p_{k+1}-p_k)s_k = \frac{1}{j}\sum_{k=1}^j s_{n_k},$$
where $n_j \le n < n_{j+1}$. It is then easy to find a subsequence of
$$x_n = s_n - \frac{1}{j}\sum_{k=1}^j s_{n_k}$$
which does not converge to $0$ in each of the cases. (In the case where $\lim s_{n_j}$ is finite, the fact that the averages $(1/j)\sum_1^j s_{n_k}$ converge to the same limit is useful.)
