Bound and limsup for cumulative sum of a random walk It is well-known from the law of iterated logarithm, that, if $X_k$ are symmetric Bernoulli random variables $\pm 1$, then $S_n= X_1 + X_2 + ... + X_n$ has this property:
$$\limsup_{n \to \infty} \frac{S_n}{\sqrt{2 n \log \log n}} = 1 \qquad \text{a.s.}$$ 
giving an order of magnitude of $n^{1/2+\varepsilon}$ for the maximum of $|S_n|$.
What bounds can be given about the cumulative sum: $$T_n = S_1 + S_2 + ... + S_n$$?
I can imagine its order of magnitude will be around $n^{3/2+\varepsilon}$, but is this quantity approached infinitely often? What is the $\limsup$?

For any $\epsilon>0$, since $|T_i-T_{i-1}| \le i$ using Azuma's inequality
\begin{align}
\sum_n P\left( \left|T_n \right|\ge \epsilon n^{\alpha}\right)
\le \sum_n 2 \exp\left( \frac{-\epsilon^2 n^{2\alpha}}{2 \sum_{i=1}^n i^2} \right)
= \sum_n 2 \exp\left( \frac{-3 \epsilon^2 n^{2\alpha-1}}{ (n+1)(2n+1)} \right)
\le  \sum_n 2 e^{-3 \epsilon^2 C n^{2\alpha-3} }
\end{align}
So, the RHS is finite if and only if if $\alpha>3/2$. If $\alpha>3/2$, then Borel-Cantelli implies that $\frac{T_n}{n^{\alpha}} \to 0$ almost surely.
If $\alpha\le 3/2$, then Borel-Cantelli does not help here. On the other hand, the $T_n$'s are not independent, so I wouldn't know what to use instead of Azuma's inequality. This may suggest that the bound $\alpha\ge 3/2+\epsilon$ is 'kind of tight', or, well, at least not that trivial!
