Tilted sum of independent random variables Let $(X_i)_i$ be a sequence of centered i.i.d. random variables with finite variance. Is it true that
$$\frac{\sum_{i=1}^{\lfloor n^{0.6} \rfloor}X_i}{\sqrt{n}}\stackrel{\mbox{a.s.}}{\longrightarrow} 0\quad ?$$
Should be a sort of corollary of the central limit theorem but I don't know how to prove the a.s. convergence. Any help?
 A: 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.
A: 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.
