Convergence of decreasing decrements of sequences

Question

Let $\beta_m\searrow 0$ such that $\alpha_m:=\beta_m-\beta_{m+1}\searrow 0$.

Define $b_n:=\inf\{m:\alpha_m<2^{-n}\}$. Is it true that $$ \sum_{n=1}^\infty \frac{b_n}{2^n}<\infty? $$

For example, if $\beta_m=\frac 1 m$, then $b_n\sim 2^{n/2}$, so that the above series converges.

A critical case is when $\beta_m=1/\log m$, whence $\alpha_m\sim 1/m(\log m)^2$, and $b_n\sim 2^n/n^2$, so the series converges.

Edited: I am sorry I had a typo: I meant $\beta_m:=1/\log m$, not $\alpha_m:=1/\log m$. In the latter case, this is a simple question. However, it is not in the former case.

Answer 1

Your critical case $ \alpha_{m} = \frac{1}{\lg(m)} $ is a counter-example.

Notice that $ b_{n} = 2^{(2^{n+1})} $, since we have

$$ \frac{1}{\lg(m)} < 2^{-n} \implies m > 2^{(2^{n})} $$

and $\alpha_{m}$ is monotonous.

Therefore,

$$ \sum_{n=1}^{\infty} \frac{2^{(2^{n+1})}}{2^{n}} = \infty $$

Actually, it is more than critical, we could make the it diverge with much less.

Answer 2

For what it's worth (this is community wiki, feel free to edit) here is the analysis of your critical case.

Consider $\beta_m = \frac{1}{\ln m}$. Then, $$\begin{align} \alpha_m &= \beta_m - \beta_{m+1} = \frac{1}{\ln m}\left( 1- \frac{1}{1 + \frac{\ln(1+\frac{1}{m})}{\ln m}}\right)= \frac{1}{\ln m}\left( 1- \frac{1}{1 + \frac{1}{m\ln m} + o(\frac{1}{m\ln m})}\right) \\ &= \frac{1}{m\ln^2 m}+ o\left(\frac{1}{m\ln^2 m}\right) \tag{1} \end{align}$$ Now, in light of the above, let's look at $$ \frac{1}{m\ln^2 m} \leq \frac{1}{2^n} \tag{2} $$ for $n\geq 1$. Rearranging and rewriting the RHS, this is equivalent to $$ 2\sqrt{m}\ln\sqrt{m} \geq 2^{n/2}\,. $$ Since $2x\ln x = 2^{n/2}$ has solution $x=\frac{2^{n/2}}{n \ln 2} + o\left(\frac{2^{n/2}}{n}\right)$ (asymptotics taken as $n\to\infty$), we get that $$ b_n \operatorname*{\sim}_{n\to\infty} \frac{2^n}{n^2\ln^2 2} $$ so that by comparison with the series $\sum_n \frac{1}{n^2\ln^2 2}$ we have $$ \sum_{n=1}^\infty \frac{b_n}{2^n} < \infty\,.\tag{3} $$

As a final remark, note that the "barrier" for that sort of $\alpha_m$ that would lead to divergence is $\alpha_m = \frac{1}{m\ln m}$. However, to get this one would need to take $\beta_m = \ln\ln m$, and this does not satisfy the assumption that $(\beta_m)_m$ be decreasing.