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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.

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The exponent, is it floor of $\sqrt n$ or round of $\sqrt n$? –  Ricbit Nov 18 '12 at 14:38
    
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1 Answer 1

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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.

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