For iid $\{X_n\}$ with finite variance,$E(X_n)=0$,does $\sum_{n=1}^\infty X_n/n^\alpha$ always converge in probability if $1/2<\alpha<1$ Suppose $\{X_n\}$ i.i.d,$E(X_n)=0$,$Var(X_n)<\infty$,for what $\alpha$ does $\sum_{n=1}^\infty X_n/n^\alpha$ converge in probability?
From strong laws of large number for iid case,if $\alpha\geq 1$ then $\sum_{n=1}^\infty X_n/n^\alpha$ converges a.s,thus converges in probability. If $X_n$ are Bernoulli random variables with parameter $1/2$,from three series theorem, $Var(\sum_{n=1}^\infty X_n/n^\alpha)=\sum_n1/n^{2\alpha}$,so $\sum_{n=1}^\infty X_n/n^\alpha$ diverges in probability if $\alpha\leq1/2$.What about the case when $1/2<\alpha<1$?Are there any counterexamples?
Update:I found Kolmogorov Strong Law of Large Numbers can be used.Take $b_n=n^{-\alpha}$,then $\sum_n Var(X_n)/b_n^2<\infty$,so $\sum_{n=1}^\infty X_n/n^\alpha$ converges a.s to 0,thus in prob.
 A: It does converge in probability for $\frac 1  2 <\alpha <1$.
Note that $\sum \frac {X_n-EX_n} {n^{\alpha}}$ converges in mean square, hence in probability. To see this compuate $E| \sum_{n=N}^{M} \frac {X_n-EX_n} {n^{\alpha}}|^{2}$ using the fact that variance of  a sum of independent r.v.'s is the sum of the variances.
A: In the case where the $X_n$ are independently Rademacher distributed, i.e. $=\pm 1$ with equal probability (hence, zero mean and finite variance), then you are dealing with a random walk with fatigue. In that case Rademacher proved that the sum $S_N=\sum_{n=1}^{N} c_n X_n$ converges with probability one if $\sum_{n=1}^{\infty} c_n^2 < \infty$ and with probability zero if $\sum_{n=1}^{\infty} c_n^2 = \infty$. This is a stronger convergence concept than you're asking for, but it gives us a strong hint.
To address the question of convergence in probability in the remaining case, suppose that the steps satisfy $\sum_{n=1}^{\infty} c_n^2 = \infty$ but $c_n\to 0$. The characteristic function (CF) of $S_N$ is given by
$$
\phi_{S_N}(t)=E\left[e^{itS_N}\right]=\prod_{n=1}^{N}E\left[e^{ic_ntX_n}\right]=\prod_{n=1}^{N}\cos\left(c_nt\right)
$$
using independence and the Rademacher CF. As $N\to\infty$, the right-hand side converges to one if $t=0$ and zero otherwise, where we use the properties of the $c_n$. (See this post for the convergence argument.) Since the CF converges pointwise to a function which is discontinuous at zero, Levy's continuity theorem tells us that $S_N$ does not converge in distribution. It follows that $S_N$ does not converge in probability.
Fatiguing random walks, as well as sums of uniform random variables, are analyzed in Morrison (1998) Random walks with decreasing steps
