Conditional convergence of $\sum \frac{(-1)^{[\sqrt{n}]}}{n^{p}}$ I need to prove that the following series:
$$
\sum \frac{(-1)^{[\sqrt{n}]}}{n^{p}}  
$$
is conditionally convergent when p=1.
Any hints would be welcome.
 A: Consider that:
$$\sum_{n=1}^{(m+1)^2-1}\frac{(-1)^{\lfloor\sqrt{n}\rfloor}}{n}=\sum_{k=1}^{m}\sum_{n=k^2}^{(k+1)^2-1}\frac{(-1)^k}{n},$$
and:
$$\frac{2k+1}{k^2+2k}\leq a_k=\left|\sum_{n=k^2}^{(k+1)^2-1}\frac{(-1)^k}{n}\right|\leq\frac{2k+1}{k^2}.$$
We have:
$$\sum_{n=1}^{(m+1)^2-1}\frac{(-1)^{\lfloor\sqrt{n}\rfloor}}{n}=\sum_{k=1}^{m}(-1)^k\, a_k,$$
and since $\left|a_k-\frac{2}{k}\right|\leq\frac{C}{k^2}$, the RHS of the last expression is a conditionally convergent series. Moreover,
$$\left|\sum_{n=1}^{M}\frac{(-1)^{\lfloor\sqrt{n}\rfloor}}{n}-\sum_{n=1}^{\lfloor\sqrt{M+1}\rfloor^2-1}\frac{(-1)^{\lfloor\sqrt{n}\rfloor}}{n}\right|\leq\frac{2\sqrt{M+1}}{M},$$
and since the RHS of the last expression converge to zero when $M$ approaches $+\infty$, the original series is conditionally convergent, too.
Here I assumed that the exponent of the $(-1)$ in the original series was the floor function of $\sqrt{n}$, but, with minor adjustments, the same argument works for the round function, too.
