Nice4

LevonNurbekian

Consider series $\sum_{n=1}^{\infty}{\frac{(-1)^{[\sqrt{n}]}}{n^{p}}.$ Find all real numbers $p$ ,such that:
a)The series above converge absolutely.
b)The series above converge but not absolutely.

**Posted:** Wed Nov 17, 2004 12:27 pm

grobber

I think this might have been posted by Harazi a long time ago.

**Posted:** Wed Nov 17, 2004 12:30 pm

Myth

a) $p>1$ ;
b) $p>1/2$ .
Am I right?

**Posted:** Wed Nov 17, 2004 12:38 pm

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Myth is out of here

grobber

Moubinool gave pretty much the same answer here.

**Posted:** Wed Nov 17, 2004 12:45 pm

LevonNurbekian

Yes,only b) $\frac{1}{2}<p\leq{1}$

**Posted:** Wed Nov 17, 2004 12:47 pm

LevonNurbekian

Guys,what about
$\sum_{n=1}^{\infty}{\frac{(-1)^{[\sqrt[q]{n}]}}{n^{p}}}$ ,where $q\in{N}.$

**Posted:** Wed Nov 17, 2004 12:53 pm

Myth

**Posted:** Wed Nov 17, 2004 1:04 pm

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Myth is out of here

LevonNurbekian

Is it too obvious?

**Posted:** Wed Nov 17, 2004 1:12 pm

Myth

I think so maybe

**Posted:** Wed Nov 17, 2004 1:23 pm

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Myth is out of here

LevonNurbekian

Anyway Myth I'm interested in your solution which would make it obvious.

**Posted:** Wed Nov 17, 2004 1:30 pm

grobber

I, for one, wouldn't call it obvious, but what's going on is easy enough to visualize (I don't think I'll be posting a full solution Smile

). Look at the case $q=2$ . We have $\sum (-1)^n\cdot\left(\frac 1{\sqrt{n^2}^{2p}}+\frac 1{\sqrt{n^2+1}^{2p}}+\ldots\frac 1{\sqrt{n^2+2n}^{2p}}+\right)$ . It's not that hard to give arguments for the fact that the term inside the brackets behaves "sort of like" $\frac 1{n^{2p-1}}$ (meaning that the ratio between the two expressions tends to a constant, in our case $2$ , I think).

Once you realize this it can't be too hard to formalize it.

**Posted:** Wed Nov 17, 2004 1:41 pm

LevonNurbekian

I agree with you grobber that it is not too obvious.Seeing that (in the case $q=2$ ) the sum inside of the brackets is like $n^{1-2p}$ is helpful in seeing that the element of the serie is not converging to $0$ in the case $p<\frac{1}{2}$ ,
but it is not sufficent to deduce that the series converge in the case $p>\frac{1}{2}$ ,because the element of the serie changes it's sign(otherwise we could say that the serie converges or not whithout any doubt).

**Posted:** Wed Nov 17, 2004 1:56 pm

Myth

There are obvious things and there are "easy to formalize" things.
This problem is a first kind thing. Wink

**Posted:** Wed Nov 17, 2004 10:11 pm

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Myth is out of here