Let me try and give an answer for $\alpha>1$ also. This one uses some technology from a
paper of mine with Boshernitzan, Kolesnik and Wierdl.

I want to use exponential sums. Using the notation of that paper we take $a(n)=n+n^\alpha$. That paper lets us control $\hat A_t(1/2)=(1/t)\sum_{n\le t}e([a(n)]/2)=(1/t)\sum_{n\le t}(-1)^{n+[n^\alpha]}$ where $e(x)=e^{2\pi i x}$ (see after Lemma 7.2).

The proofs of Theorem 3.4 and Theorem 7.1 give (if you read carefully) the existence of an $\epsilon>0$ and a $C$ such that $|\hat A_t(1/2)| < Ct^{-\epsilon}$ for all $t$. This says that the difference between the number of $+1$'s and the number of $-1$'s (the *discrepancy*) for $n\le t$ is at most $t^{1-\epsilon}$ (ignoring constants from now on).

Now let $K>2/\epsilon$ and divide the integers into blocks $I_j=(j^K,(j+1)^K]$. The discrepancy up to $j^K$ is at most $j^{K-2}$ by the above. Similarly the discrepancy up to $(j+1)^K$ is also at most $j^{K-2}$. So the discrepancy in the $I_j$ block is at most $j^{K-2}$.

We now have $\sum_{n\in I_j}(-1)^{n+[n^\alpha]}/n = \sum_{n\in I_j}(-1)^{n+[n^\alpha]}/j^K + \sum_{n\in I_j}(-1)^{n+[n^\alpha]}(1/n-1/j^K)$.

By the above comment, the first term contributes $j^{-2}$. In the second term, there are $j^{K-1}$ terms, each contributing in absolute value at most $j^{-K-1}$ giving a maximum total contribution of $j^{-2}$.

It follows that the contributions from the $I_j$-blocks are absolutely summable and we're done.

Of course $\alpha<0$ is trivial so this is good for all $\alpha$ except the positive integers.

integrals of special functions! – Wadim Zudilin Jan 13 '11 at 9:43