1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.