How find series $\sum_{n=1}^{\infty}\frac{(-1)^{[\sqrt[m]{n}]}}{n^a}$? 
let $m$ is give a positive integers,
  Determine for which values of $a$,the series $$\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt[m]{n}]}}{n^a}$$ converges  

where $[x]$ is the largest integer not greater than $x$.
and  I have see this not hard problem :show that
$$\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt{n}]}}{n}$$ converges,
solution: note

$$\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt{n}]}}{n}=\sum_{n=1}^{\infty}(-1)^n\left(\dfrac{1}{n^2}+\dfrac{1}{n^2+1}+\cdots+\dfrac{1}{n^2+2n}\right)$$
  and we have
  let $$a_{n}=\dfrac{1}{n^2}+\dfrac{1}{n^2+1}+\cdots+\dfrac{1}{n^2+2n}$$
  $$0<a_{n}<\dfrac{2n+1}{n^2}<\dfrac{3}{n}\longrightarrow 0$$
  and
  $$a_{n}-a_{n+1}=\dfrac{1}{n^2}-\dfrac{1}{(n+1)^2}+\dfrac{1}{n^2+1}-\dfrac{1}{(n+1)^2+1}+\cdots+\dfrac{1}{n^2+2n}-\dfrac{1}{(n+1)^2+2n}-\left(\dfrac{1}{(n+1)^2+2n+1}+\dfrac{1}{(n+1)^2+2(n+1)}\right)$$
  so
  $$a_{n}-a_{n+1}=(2n+1)\left(\dfrac{1}{n^2(n+1)^2}+\cdots+\dfrac{1}{(n^2+2n)(n^2+4n+1)}\right)-\dfrac{1}{n^2+4n+2}-\dfrac{1}{n^2+4n+3}$$
  so
  $$a_{n}-a_{n+1}>\dfrac{(2n+1)^2}{(n^2+2n)(n^2+4n+1)}-\dfrac{2}{n^2+4n+1}=\dfrac{2n^2+1}{(n^2+2n)(n^2+4n+1)}>0$$
  so
  $$\sum_{n=1}^{\infty}(-1)^na_{n}=\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt{n}]}}{n}$$ converges

But for my problem,I can't 
My try:
$$\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt[m]{n}]}}{n^a}=\sum_{n=1}^{\infty}(-1)^n\sum_{k=n^m}^{(n+1)^m-1}\dfrac{1}{k^a}$$
let
$$b_{n}=\sum_{k=n^m}^{(n+1)^m-1}\dfrac{1}{k^a}$$
then how prove 
$$b_{n}\longrightarrow 0$$and $$b_{n}>b_{n+1}$$

Let's start with the observation that if $a > 1$, then the sum converges absolutely, and if $a \leqslant 0$, the terms of the sum don't converge to $0$, hence the sum cannot converge. So in the following, we always assume $0 < a$.
The case $m = 1$ is immediate by Leibniz' criterion, $\sum\limits_{n=1}^\infty \dfrac{(-1)^n}{n^a}$ converges if and only if $a > 0$.
Generally, as you found, we must consider the segments
$$b_n = \sum_{k=n^m}^{(n+1)^m-1} \frac{1}{k^a}.$$
We do need that $b_n \to 0$, but we don't necessarily need that $b_n > b_{n+1}$. If we can write $b_n = c_n + d_n$, where $c_n \downarrow 0$, and $\sum_n d_n$ converges absolutely, that suffices to show the (conditional) convergence of the original sum. Now,
$$b_n = \underbrace{\int_{n^m}^{(n+1)^m} \frac{dt}{t^a}}_{c_n} + \underbrace{\sum_{k=n^m}^{(n+1)^m-1} \int_{k}^{k+1}\left(\frac{1}{k^a} - \frac{1}{t^a}\right)\,dt}_{d_n}.$$
We have $\frac{1}{k^a} - \frac{1}{t^a} = \frac{a(t-a)}{\xi^{a+1}}$ for some $\xi \in [k,t]$ by the mean value theorem, so
$$0 < \int_k^{k+1} \frac{1}{k^a} - \frac{1}{t^a}\,dt < \frac{a}{2k^{a+1}},$$
whence $$\sum_{n=1}^\infty \lvert d_n\rvert < \frac{a}{2}\sum_{k=1}^\infty \frac{1}{k^{a+1}} < \infty.$$
It remains to consider $c_n$. For $a = 1$, we get
$$c_n = \int_{n^m}^{(n+1)^m} \frac{dt}{t} = \log \left((n+1)^m\right) - \log \left(n^m\right) = m\left( \log (n+1)-\log n\right) = m\log\left(1+\frac1n\right).$$
It's evident that $c_n \to 0$ and $c_n > c_{n+1}$ in this case.
For $0 < a < 1$, we get
$$c_n = \int_{n^m}^{(n+1)^m} \frac{dt}{t^a} \overset{t=u^m}{=} \int_n^{n+1} \frac{mu^{m-1}}{u^{ma}}\,du = m\int_n^{n+1} u^{m(1-a)-1}\,du,$$
and we see that $c_n \to 0$ if and only if $m(1-a)-1 < 0 \iff a > \dfrac{m-1}{m}$. $c_n$ is easily seen to be monotonic (constant for $m(1-a)-1 = 0$, increasing for $m(1-a)-1 > 0$, and decreasing in the convergent case).
Wrapping up:
$$\sum_{n=1}^\infty \frac{(-1)^{\lfloor \sqrt[m]{n}\rfloor}}{n^a}$$
converges if and only if $a > \dfrac{m-1}{m}$.
