Evaluate $\sum_{n=1}^\infty \frac{(-1)^{[\sqrt{n}]}}{n^p}$ for $ 0 < p \leq 1$ How do you evaluate $$\sum_{n=1}^\infty \frac{(-1)^{[\sqrt{n}]}}{n^p}$$ for the case $0 < p \leq 1$?
Now I have successfully proved that the series converges for the case $p > 1$ and divergence of $p\leq 0$ is trivial, but I cannot deal with the case $0 < p \leq 1$. Wolfram Alpha tells me that it actually diverges but I completely have no idea on how to prove that. 
Here $[\sqrt{n}]$ stands for the largest integer that is smaller than $\sqrt{n}$.

$$S(N) = \sum_{n=1}^{N}(-1)^{\lfloor\sqrt{n}\rfloor}$$
is such that $S((2M)^2)=-2M$ and $S((2M+1)^2)=2M-1$. 
In particular $|S(N)|\leq\sqrt{N}$ and this inequality is sharp for any square $N$, but the sign of $S(N)$ depends on the parity of $\lfloor\sqrt{N}\rfloor$. If we assume $p\in\left(\frac{1}{2},1\right]$ and apply summation by parts we have that the original series behaves like 
$$ \sum_{n\geq 1}\frac{S(n)}{n^{p+1}}$$
which is absolutely convergent. Summation by parts also gives that the original series is not convergent for $p\in\left(0,\frac{1}{2}\right)$. It follows that the only non-trivial case to be discussed is $p=\frac{1}{2}$. In this case the partial sums of the original series keep oscillating between two finite values and the original series is not convergent. Summarizing, Kronecker's lemma or something equivalent gives that convergence happens for $p>\frac{1}{2}$.
