# 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.

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.)