Tilted sum of independent random variables

By looking at the cases where $n$ is in an interval of the form $\left[N^{1/0.6},(N+1)^{1/0.6}\right)$, we notice that the wanted convergence holds if we manage to prove the following: for each positive $\eta$, the following convergence holds almost surely: $$ \lim_{n\to +\infty}\frac 1{n^{1/2+\eta}}\left\lvert \sum_{i=1}^nX_i \right\rvert=0. $$ One could can use the bounded law of the iterated logarithms, which says that under the conditions of the opening post, the random variable $$ M:=\sup_{n\geqslant 3}\frac 1{\sqrt{n\log\log n}}\left\lvert \sum_{i=1}^nX_i \right\rvert $$ is almost surely finite. Therefore, $$ \frac 1{n^{1/2+\eta}}\left\lvert \sum_{i=1}^nX_i \right\rvert\leqslant \frac{\sqrt{\log\log n}}{n^\eta}M. $$

An other way is to control the moments of order two of $2^{-N\left(1/2+\eta\right)}\max_{1\leqslant n\leqslant 2^N}\left\lvert \sum_{i=1}^nX_i \right\rvert$ by using Doob's inequality. This will prove finiteness of $$ \sum_{N\geqslant 1}2^{-N\left(1/2+\eta\right)}\max_{1\leqslant n\leqslant 2^N}\left\lvert \sum_{i=1}^nX_i \right\rvert, $$ which is sufficient to conclude.

Let $\alpha\in (0,1)$ and $m=\lfloor n^{\alpha} \rfloor$. Then $$ \frac{1}{\sqrt{n}}\left|\sum_{i=1}^{\lfloor n^{\alpha} \rfloor}X_i\right|\le\frac{1}{m^{1/(2\alpha)}}\left|\sum_{i=1}^{m}X_i\right|\to 0 \quad\text{a.s.} $$ by the Marcinkiewicz–Zygmund SLLN.