Investigate on the convergence of the series, with parameter $$\sum_{n=1}^\infty (-1)^{[n^a]}n^{-a}$$ 
Learn for what $a \in \mathbb {R}$ sequence converges. Symbol [x] means integer part (floor?) of number x.
It's my first question, if you will haveany complaints - write about it please.
 A: This series converges if and only if $a > \frac12$.
Let's deal with the easy parts first:


*

*It is absolutely convergent for $a > 1$, this is well known.

*It diverges for $a \le 0$, as the terms do not go to zero.

*Now, for the important part, $0 < a \le 1$, the idea is to group the terms together for which $\lfloor n^a\rfloor$ is the same:
$$\sum_{n=1}^{\infty} (-1)^{\lfloor n^a\rfloor}n^{-a} = \sum_{k=1}^\infty (-1)^k x_k \text{ where } x_k = \sum_{\lfloor n^a\rfloor=k} n^{-a} = \sum_{k^{1/a} \le n < (k+1)^{1/a}} n^{-a}$$
So, the original series is equivalent to a sign-alternating series: $\sum_{k=1}^\infty (-1)^k x_k$. Let's look at $x_k$ more closely:
$$ x_k = \sum_{\lfloor n^a\rfloor=k} n^{-a} = \sum_{\lfloor n^a\rfloor=k} \left(\frac{1}{k} + O(\frac1{k^2})\right) = \left((k+1)^{1/a} - k^{1/a} + O(1)\right)\left(\frac{1}{k} + O(\frac1{k^2})\right) = k^{1/a}\left(\frac1{ak} + O(\frac1{k^2})\right)\left(\frac{1}{k} + O(\frac1{k^2})\right) = \frac{k^{\frac1a - 2}}{a} + O(k^{\frac1a -3}) $$
From which we can easily see that:


*If $a \le \frac12$ then $\frac1a -2 \ge 0$ and $\lim x_k \ne 0$, so the series diverges.

*If $a > \frac 12$ then $\frac1a -2 < 0$ and $\frac1a -3 < -1$, so
$$ \sum_{k=1}^\infty (-1)^k x_k = \frac1a \sum_{k=1}^\infty (-1)^kk^{\frac1a -2} + \sum_k O(k^{\frac1a -3})$$
Because the second part absolutely converges. And the first part converges because it's a sign-alternating series with terms going to zero monotonously.
A: A quick answer showing divergence for $a=1/2$.
For $N\geq 1$, on can write the partial sum $S_{N^2}$ as
$$S_{N^2} = \sum_{n=1}^{N^2} \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{\sqrt{n}} = \sum_{k=1}^N (-1)^k \sum_{\ell=k^2}^{(k+1)^2-1} \frac{1}{\sqrt{\ell}}\,.$$
Now, since
$$\sum_{\ell=k^2}^{(k+1)^2-1} \frac{1}{\sqrt{\ell}} = 2 + \frac{1}{2k^2} + o\left(\frac{1}{k^2}\right)$$ when $k\to\infty$, and recalling that the series $\sum_{k=1}^\infty (-1)^k\left(\frac{1}{2k^2} + o\left(\frac{1}{k^2}\right)\right)$ is absolutely convergent, we get that $(S_{N^2})_N$ converges is, and only if, $2\sum_{k=1}^n (-1)^k$ does. And it does not.

Here is a plot illustrating the (lack of) convergence, for $a=1/2$ (for partial sums $S_n$ up to $n=2000$).

