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$ \sum_{n=1}^{\infty}\frac{\left(-1\right)^{\lfloor\sqrt{n+1}\rfloor}-\left(-1\right)^{\lfloor\sqrt{n}\rfloor}}{n} $

Im having trouble trying to determine if this series converge absolutely or conditionally. Any ideas will help. I noticed that when $ \sqrt{n+1} \in Z $ then $ \left(-1\right)^{\lfloor\sqrt{n+1}\rfloor}-\left(-1\right)^{\lfloor\sqrt{n}\rfloor} =-2 $ or $ \left(-1\right)^{\lfloor\sqrt{n+1}\rfloor}-\left(-1\right)^{\lfloor\sqrt{n}\rfloor}=2 $

But i cant tell more then that.

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  • $\begingroup$ I think you can tell more than that since by what you already wrote it is clear the series cannot converge absolutely... $\endgroup$
    – DonAntonio
    Commented Apr 17, 2020 at 22:49
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    $\begingroup$ @DonAntonio: yes, it can, because most of the terms are zero. The surviving ones decrease fast enough so the sum converges absolutely. $\endgroup$
    – Ross Millikan
    Commented Apr 17, 2020 at 22:51
  • $\begingroup$ @RossMillikan I'm referring to what the OP wrote: that the numerator is either $\;2\;$ or $\;-2\;$ ... $\endgroup$
    – DonAntonio
    Commented Apr 17, 2020 at 23:09
  • $\begingroup$ @DonAntonio: only when $n+1$ is a perfect square, which OP represented as $\sqrt {n+1} \in \Bbb Z$, implying it is zero other times. $\endgroup$
    – Ross Millikan
    Commented Apr 17, 2020 at 23:13

1 Answer 1

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You need to turn what you noticed, which is important, into the fact that the only surviving terms have $n+1=k^2$, so they have $k^2-1$ in the denominator. Now you can use the fact that the sum of inverse squares converges.

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  • $\begingroup$ Im having trouble writing formal proof. I want to write this: $ \sum_{n=1}^{\infty}\frac{|\left(-1\right)^{\lfloor\sqrt{n+1}\rfloor}-\left(-1\right)^{\lfloor\sqrt{n}\rfloor}|}{n}=\sum_{n=1}^{\infty}\frac{|\left(-1\right)^{\lfloor\sqrt{n+1}\rfloor}-\left(-1\right)^{\lfloor\sqrt{n}\rfloor}|}{k^{2}-1}\leq\sum_{n=1}^{\infty}\frac{2}{k^{2}} $ but im not sure how to do it correct. because in the last sum we dont have dependence on n in the term so it looks wrong. (k is the number such $ k^2=\sqrt{n+1} $ $\endgroup$
    – FreeZe
    Commented Apr 17, 2020 at 23:53
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    $\begingroup$ I would change the sum variable to $k$. To do that you have to argue that all the terms you lose because they do not correspond to a whole number $k$ are zero. You also want to start with $k=2$ to avoid the division by $0$ which is OK because you did not have $n=0$ which would cause $k=1$. $\endgroup$
    – Ross Millikan
    Commented Apr 18, 2020 at 0:09

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