Convergence of decreasing decrements of sequences 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.
 A: 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.
A: 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.
