4
$\begingroup$

Main question: Let $a_n \geq 0$ and $b_n = \sum_{k=1}^n a_k \uparrow \infty$. Is it true that $$ \sum_{n=1}^\infty \frac{a_n}{b_n^2} < \infty \,? $$

I strongly suspect there is either a short, slick proof that I am not seeing or a simple counterexample I've missed.

Some motivation. Let $X_1,X_2,\ldots$ be independent Poisson random variables where $X_n$ has mean $a_n$ and $S_n = X_1 + \cdots + X_n$. If $\sum_n a_n = \infty$ then it's trivial to verify that $S_n/b_n \to 1$ in $L_2$ and so, also, in probability. I am quite sure that this result should also hold almost surely. ($S_n$ is Poisson with a diverging mean.) One way to go about that would be to use Kolmogorov's convergence criterion in concert with Kronecker's lemma so that if $$ \sum_{n=1}^\infty \text{var}(S_n/b_n) = \sum_{n=1}^\infty \frac{a_n}{b_n^2} < \infty $$ then almost-sure convergence would follow.

Trivial bounds don't work, but there is some hope. Clearly $a_n / b_n \leq 1$ uniformly in $n$, but this crude bounding is not enough to assert the desired result. For example, taking $a_n = 1$ shows that $\sum_n a_n/b_n^2 < \infty$, but, of course, $\sum_n b_n^{-1} = \infty$. So, we need something stronger than this.

On the other hand, let $a_n = (n \log n)^{-1}$, which is just on the boundary of divergence, i.e., $\sum_n a_n = \infty$ as needed, but if we replaced the exponent with $-(1+\delta)$ for any $\delta > 0$, then it would converge. Yet, in the $a_n = (n\log n)^{-1}$ case we have $$ b_n \approx \int_e^n \frac{\mathrm d x}{x \log x} = \log \log n $$ so that $$ \sum_n \frac{a_n}{b_n^2} \leq C \sum_n \frac{1}{n \log n (\log \log n)^2} < \infty \,. $$

This same basic argument works if we introduce more iterated log terms in the definition of $a_n$. Furthermore, it's clear that if $\sum_n a_n < \infty$, then $\sum_n a_n/b_n^2 < \infty$.

And, that brings me back to the main question.

$\endgroup$

1 Answer 1

3
$\begingroup$

Hint for Main Question: These formulas hold: $$\frac{a_n}{b_n^2} =\frac{b_n-b_{n-1}}{b_n^2}$$ and $$2\le \frac{b_n}{b_{n-1}}+\frac{b_{n-1}}{b_n}.$$ So we can get following inequailty: $$\frac{a_n}{b_n^2}\le \frac{1}{b_{n-1}}-\frac{1}{b_n}.$$ Use this inequality to take upper bound for this sum.

$\endgroup$
1
  • 1
    $\begingroup$ +1, thanks. I had used the first expression but had a hard time getting anywhere. Note We can actually get away with something even cheaper. Since $b_n \geq b_{n-1}$, then $$\sum_n \frac{b_n-b_{n-1}}{b_n^2} \leq \sum_n \frac{b_n - b_{n-1}}{b_n b_{n-1}} = \sum_n (b_{n-1}^{-1} - b_n^{-1}) \,.$$ In other words, your middle expression is not needed. $\endgroup$
    – bpt
    Jan 17, 2013 at 13:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .