I was going through Peter Morter's and Yuval Peres' textbook Brownian Motion, and early on in the book they provide a proof for the following statement:
There exists a constant $C>0$ such that, almost surely, for every sufficiently small $h>0$ and all $0\leq t\leq 1-h$, $$|B(t+h) - B(t)|\leq C\sqrt{h\log(1/h)}$$
I understand every line of the provided proof save the last line, and outline it here: they first define $B(t) = \sum_{i=0}^{\infty}F_n(t)$, where $F_n(\cdot)$ are a sequence of piecewise linear functions defined such that they interpolate values at the dyadic rationals. They then establish that for some $n>N$, $$\lVert F_n'\rVert_\infty \leq \frac{\lVert F_n\rVert_\infty}{2^{-n}}\leq c\sqrt{n}2^{n/2}$$ They then establish a bound on $|B(t+h) - B(t)|$ using this, and through using the mean value theorem and some triangle inequality, the bound winds up looking like: $$|B(t+h) - B(t)|\leq h\sum_{n=0}^{N}\lVert F_n'\rVert_\infty + ch\sum_{n=N+1}^{\ell}\sqrt{n}2^{n/2} + 2c\sum_{n=\ell+1}^{\infty}\sqrt{n}2^{-n/2}$$ for $t, t+h\in[0,1]$ and $\ell>N$.
The authors then say something that confuses me: they claim that for $\ell>N$ that is defined such that $2^{-\ell} < h \leq 2^{-\ell+1}$, that the second and third summands become bounded by $\sqrt{h\log(1/h)}$. Maybe I'm missing something here, but I don't see why that is the case. Can someone provide an explanation as to why this final step is justified?
For the second term (which I assume should have summands $\sqrt{n} 2^{-n/2}$), we can bound the $\sqrt{n}$ term by $\sqrt{\ell}$, and what remains can be bounded by the geometric series $\sum_{n = N+1}^{\infty} 2^{-n/2}$, a constant depending on $N$ (which we fix). For $h < 1$, we also have $h < \sqrt{h}$. So the second term is $O(\sqrt{h \ell}) = O(\sqrt{h \log(1/h)}$ since $h \leq 2^{-\ell + 1}$ means $2^{\ell - 1} \leq 1/h$ and in turn $\ell = O(\log(1/h))$.
For the third term, $\sum_{n = \ell+1}^\infty \sqrt{n} 2^{-n/2} = \sqrt{\ell + 1} 2^{-(\ell + 1) / 2} \sum_{n=0}^{\infty} \sqrt{1 + n / (\ell + 1)} 2^{-n/2}$. Note that we can bound the geometric series by $\sum_{n=0}^{\infty} \sqrt{1 + n/(\ell + 1)} 2^{-n/2} \leq \sum_{n = 0}^\infty n 2^{-n/2}$ (with appropriate choice of $N$), and the latter is bounded by a constant again (recall that $x + 2x^2 + 3x^3 + \dots = x / (1 - x)^2$ for $|x| < 1$). So what's left in front is $\sqrt{\ell + 1} = O(\sqrt{\log(1/h)})$ and $\sqrt{2^{-(\ell + 1)}} = O(\sqrt{h})$ again.