$ %Define large versions of lfloor and rfloor (big enough to handle sqrt). \newcommand{\lf}{\large \lfloor \normalsize} \newcommand{\rf}{\large \rfloor \normalsize} $ Note: $\lf\sqrt{n}\rf$ denotes integer part of $\sqrt{n}$ (floor rounding).

So, I have such series $$A :=\sum_{n=1}^{\infty} a_n$$

where $$a_n = \frac{(-1)^{\lf\sqrt{n}\rf}}{n^{\alpha}}$$

This is what I tried to do:

Let's group our $a_n$ such way that by doing that all elements in group will have same sign. If series that we get after grouping sequence such way converges, initial series will converge also.

Let's denote by $t$ minimal $n$ such that $\lf \sqrt{n} \rf = t$.  But that means that $n = t^2$. Now let's find other elements of that group.

$ \lf\sqrt{n}\rf = \lf\sqrt{t^2}\rf = \lf\sqrt{t^2 + 1}\rf = \lf\sqrt{t^2 + 2}\rf = \cdots = \lf\sqrt{t^2 + 2t}\rf$.

We have $2t + 1$ elements in each group.

So we can see how our series would look like

$$\large\sum_{t = 1}^{\infty}\normalsize \sum_{n = t^2}^{t^2 + 2t}\frac{(-1)^t}{n^{\alpha}}$$

Now we can see that

$$(2t + 1) \cdot \frac{1}{(t^2 + 2t)^{\alpha}} \le \sum_{n = t^2}^{t^2 + 2t}\frac{1}{n^{\alpha}} \le (2t + 1) \cdot \frac{1}{(t^2)^{\alpha}}$$

So, is it accurate to say that for $\alpha \in (\frac{1}{2}, 1)$ we have convergence here? For instance, by applying Leibniz criterion. I have doubts since I apply it not to my sequence $a_n$ but to the sequence that I was able to use to evaluate it from right side (however, I believe that the one on the left is also such that $b_n \searrow 0$ and $b_n \ge 0$ if $\alpha \in \big(\frac{1}{2}, 1)$ ).

And of course for $\alpha = 0$ it diverges.

But what about case of $\alpha \in (0, \frac{1}{2}\big]$? How to see what happens there? And did I show everything right in case of $\alpha \in \big(\frac{1}{2}, 1)$?

Idea: may be we can use evaluation of sum of each of block using definite integral? At least it’s absolute value. $|\sum_{n = t^2}^{t^2 + 2t}\frac{(-1)^t}{n^{\alpha}} - (-1)^t\int_{t^2}^{t^2 + 2t + 1} x^{-\alpha} dx| \le (t^2)^{-\alpha} - (t^2+2t + 1)^{-\alpha}$. I tried to use it but it did not help much.

All help will be appreciated!

  • $\begingroup$ When you say “denote by $t$ minimal $n$ such that $\lfloor n \rfloor = t$”,  do you mean “such that $\lfloor \sqrt{n} \rfloor = t$”? $\endgroup$
    – Scott
    May 23 at 20:25
  • $\begingroup$ @Scott Yes, I do. Edited $\endgroup$ May 23 at 20:37

Both the upper and lower bound for $\sum_{n=t^2}^{t^2 +2t} \frac{1}{n^{\alpha}}$ are asymptotic to $$\frac{2}{t^{2\alpha-1}} $$ Now, by the Alternating Series Test, $\sum(-1)^t\frac{2}{t^{2\alpha -1} } $ will converge when $2\alpha -1 \gt 0 \iff \alpha \gt \frac 12$ whereas if $2\alpha -1 \le 0 \iff \alpha \le \frac 12$, the modulus of each term is $\ge 2$ and so there is divergence.

  • $\begingroup$ Does it mean that I can't say anything about convergence of initial sequence? $\endgroup$ May 23 at 18:47
  • $\begingroup$ Did you forget the factor $(-1)^t$? $\endgroup$
    – Mark Viola
    May 23 at 18:49
  • $\begingroup$ @math-traveler No, I’ve changed my answer. $\endgroup$
    – Tavish
    May 23 at 18:59
  • $\begingroup$ @MarkViola It somehow slipped past my eyes. $\endgroup$
    – Tavish
    May 23 at 18:59
  • 1
    $\begingroup$ @Tavish Are you sure you can do that even though the sequences are not positive? $\endgroup$
    – Axel
    May 23 at 19:13

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