I owe the idea of asking this question to Max Muller and
[his curiosity](https://mathoverflow.net/questions/26035/).


*What is the set of $\alpha$ in the interval $0\le\alpha < 1$ for which
the alternating sum*
$$
\sum\_{n=1}^\infty\frac{(-1)^{n+[n^\alpha]}}n
$$
*converges*? Here $[\ \cdot\ ]$ denotes the integral part of a number.


It clearly converges when $\alpha=0$ and
[my post](https://mathoverflow.net/questions/26035/evaluation-of-the-following-series-s-1-2-times3-1-5-times6-1-7-tim/48433#48433)
to Max's answer implies the convergence when $\alpha=1/2$.
What about more general $\alpha$? Of course, the question is meaningful
for positive $\alpha\notin \mathbb Z$ as well, but then it seems to be much harder.



Convergence of alternating harmonic sums
It converges for all $0\le\alpha<1$. Define the $k$-*block* to be the set of $n$ such that $[n^\alpha]=k$ (it ranges from $\lceil k^{1/\alpha}\rceil$ to $\lceil (k+1)^{1/\alpha}\rceil-1$). 


The absolute value of the contribution to the sum from the $k$-block is at most the reciprocal of its left endpoint (the terms form an alternating series) - that is approximately $k^{-1/\alpha}$. Hence the contributions from the $k$-blocks are absolutely summable. Bingo


