I'm stuck on details of the following proof.
here $S_n$ is the value of a simple symmetric random walk after $n$ time steps, and $b_n=[2n\log\log n]^{1/2}$. I don't see how 'The result follows from 5.2.4 and Borel-Cantelli on letting $\alpha\rightarrow 1$. I tried to show it but couldn't.
Also, I don't understand what the second highlighted inequalities are saying, or how we get them. In particular, I don't know what $\gamma(m)$ is. I know that they sometimes use $\gamma$ to denote the $\lim\sup$ in the statement of the lemma, but I don't think that's how they're using it.
These are from p. 13 of the following lecture notes http://www.math.lsa.umich.edu/~conlon/math625/chapter1.pdf.
1) You know from Borel-Cantelli that if the above sum is finite, then only finitely many of those events happen, i.e. there will be some finite $n_0$ after which for any $n > \alpha^{n_0}$, the sequence $\frac{S_n}{b_n}$ will be less than $\alpha$. Since this holds true for any $\alpha > 1$, you can assume that the lim sup is bounded by 1.
2) There, they are just using $\gamma(m)$ as a function that tends to 1, and it isn't really necessary for the proof; they are just trying to convince you that each of these intervals is essentially a constant so that you are not really introducing slack when you look at the intervals of the form $[\alpha^n, \alpha^{n+1}]$ (as opposed to looking at each $b_n$ individually).
If you want the explicit form of that function, the following calculations should help:
Since you know that $\alpha^m < r, k < \alpha^{m+1}$, and $b_n$ is monotone in $n$, note that $b_r \leq b_{\alpha^{m+1}}, k \geq b_{\alpha^m}$.
Then, $\frac{b_r}{b_k} \leq \frac{(2 \alpha^{m+1} \log \log \alpha^{m+1})^{1/2}}{(2\alpha^m \log \log \alpha^m)^{1/2}} = \sqrt{\alpha} \frac{\log(m+1) + \log\alpha}{\log(m) + \log\alpha}$, and now you can denote $\gamma(m) = \frac{\log(m) + \log\alpha}{\log(m+1) + \log\alpha}$.