Konvergenz einer Reihe zeigen

13.04.2013, 13:57

lurchi_der_lurch

Konvergenz einer Reihe zeigen

Meine Frage:
Hallo, ich hab folgende Reihe mit $\begin{eqnarray*} D,\mu>0 \end{eqnarray*}$ :

$\begin{eqnarray*} \sum\limits_{k\in\mathbb{Z}^n}\frac{D}{(1+\lvert k\rvert)^{n+\mu}} \end{eqnarray*}$ .

Ich soll zeigen, dass diese Reihe konvergiert.

Meine Ideen:
Gibt's ein mehrdimensionales Integralvergleichskriterium?

13.04.2013, 15:26

Lurchi_der_Lurch

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hello i decided to registrate here so welcome to everybody here
unfortunately i am not able to ask my questions in german
therefore i have to ask in english although its not my mother language as well
i try to use german as often as possible
----
anyhow:
do you have an idea for my question above? i am still confused

i did not find a multidimensional integral criteria (i am only familiar with the one dimensional formulation) but if such a variation exists i think one can show the convergence with that?

greetings

13.04.2013, 15:40

Che Netzer

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By $\begin{eqnarray*} \lvert k\rvert \end{eqnarray*}$ you probably mean $\begin{eqnarray*} \lvert k_1\rvert+\dotsb+\lvert k_n\rvert \end{eqnarray*}$ .
If so, let $\begin{eqnarray*} m\in\mathbb N \end{eqnarray*}$ and use that the number of $\begin{eqnarray*} k\in\mathbb Z^n \end{eqnarray*}$ with $\begin{eqnarray*} \lvert k\rvert=m \end{eqnarray*}$ is polynomially increasing in $\begin{eqnarray*} m \end{eqnarray*}$ with degree $\begin{eqnarray*} n \end{eqnarray*}$ .
By that you may deduce the convergence from the convergence of some "one dimensional series".

13.04.2013, 15:45

Lurchi_der_Lurch

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thank you for answering

to be honest i do not know what is meant with $\begin{eqnarray*} \lvert k\rvert \end{eqnarray*}$ .

my thought was that it is the euclidian norm, i.e.

$\begin{eqnarray*} \lvert k\rvert=\sqrt{k_1^2+\hdots k_n^2} \end{eqnarray*}$ , whereas

$\begin{eqnarray*} k=(k_1,\hdots,k_n) \end{eqnarray*}$

is that possible?

13.04.2013, 15:51

Che Netzer

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Yes, that could be possible, too.
Anyway, both norms are equivalent, so it doesn't matter Augenzwinkern

13.04.2013, 15:55

Lurchi_der_Lurch

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right - -

but i do not see your point yet [it is so long ago...]

what do you mean with

"[...] deduce the convergence from the convergence of some "one dimensional series"?

maybe [if you have time] you can detail it for me?

edit:

do you mean

$\begin{eqnarray*} \sum\limits_{k\in\mathbb{Z}^n}\frac{C}{(1+\lvert k\rvert)^{n+\epsilon}}=\sum\limits_{m\in\mathbb{N}}\frac{C}{(1+m)^{n+\varepsilon}} \end{eqnarray*}$ ?

but this cannot be right because the right side does not take into consideration how often the sum m appears by different k's.

13.04.2013, 16:33

Lurchi_der_Lurch

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for example when $\begin{eqnarray*} n=1 \end{eqnarray*}$ it is imo

$\begin{eqnarray*} \sum\limits_{k\in\mathbb{Z}}\frac{C}{(1+\lvert k\rvert)^{1+\epsilon}}=C+2\cdot\sum\limits_{m\in\mathbb{N}}\frac{C}{(1+m)^{1+\epsilon}} \end{eqnarray*}$

but i do not know what it is in common for n

13.04.2013, 17:03

Che Netzer

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Zitat:

Original von Lurchi_der_Lurch
do you mean

$\begin{eqnarray*} \sum\limits_{k\in\mathbb{Z}^n}\frac{C}{(1+\lvert k\rvert)^{n+\epsilon}}=\sum\limits_{m\in\mathbb{N}}\frac{C}{(1+m)^{n+\varepsilon}} \end{eqnarray*}$ ?

Almost. There is some factor (the number of $\begin{eqnarray*} k\in\mathbb Z^n \end{eqnarray*}$ with $\begin{eqnarray*} \lvert k\rvert=m \end{eqnarray*}$ ) missing which increases in $\begin{eqnarray*} O(m^{n-1}) \end{eqnarray*}$ .

13.04.2013, 17:05

Lurchi_der_Lurch

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yes i thought so

but i do not know how to determine this factor and where to put it in the right sum

13.04.2013, 17:09

Che Netzer

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Actually, the exact factor doesn't matter.
Instead, we can estimate it by $\begin{eqnarray*} \leq\tilde Cm^{n-1} \end{eqnarray*}$ for almost all $\begin{eqnarray*} m\in\mathbb N \end{eqnarray*}$ and some $\begin{eqnarray*} \tilde C>0 \end{eqnarray*}$ .

13.04.2013, 17:23

Lurchi_der_Lurch

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sorry, i do not see why the number of the k with $\begin{eqnarray*} \lvert k\rvert=m \end{eqnarray*}$ increases in $\begin{eqnarray*} \mathcal{O}(m^{n-1}) \end{eqnarray*}$ .

13.04.2013, 19:19

Lurchi_der_Lurch

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hey again!

i thought about it and came to the following

let $\begin{eqnarray*} g(m):=\mbox{card}\left\{k\right\} \end{eqnarray*}$ be the number of those $\begin{eqnarray*} k=(k_1,\hdots,k_n)\in\mathbb{Z}^n \end{eqnarray*}$ which fullfill $\begin{eqnarray*} \lvert k\rvert=\lvert k_1\rvert+\hdots+\lvert k_n\rvert=m, m\in\mathbb{N} \end{eqnarray*}$

then f.a. $\begin{eqnarray*} m\in\mathbb{N} \end{eqnarray*}$

$\begin{eqnarray*} \sum\limits_{k\in\mathbb{Z}^n}\frac{C}{(1+\lvert k\rvert)^{n+\epsilon}}=C+\sum\limits_{m\geq 1}g(m)\cdot\frac{C}{(1+m)^{n+\epsilon}}\leq C+\sum\limits_{m\geq 1}\frac{g(m)}{m^{n+\epsilon}}\cdot C\leq C+\sum\limits_{m\geq 1}\frac{g(m)}{m^{n}}\cdot C\leq C+\sum\limits_{m\geq 1}\frac{g(m)}{m^{n-1}}\cdot C \end{eqnarray*}$

is that okay? and what is the next step in ordner to reach the aim?

with regards

Lurchi

13.04.2013, 19:34

Che Netzer

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Let's just assume that $\begin{eqnarray*} g(m)\leq\tilde Cm^{n-1} \end{eqnarray*}$ for almost every $\begin{eqnarray*} m\in\mathbb N \end{eqnarray*}$ and some $\begin{eqnarray*} \tilde C>0 \end{eqnarray*}$ Augenzwinkern

Maybe the following thought helps: Consider the set $\begin{eqnarray*} \{k\in\mathbb Z^n:\lvert k\rvert=m\} \end{eqnarray*}$ as some kind of "dotted $\begin{eqnarray*} n \end{eqnarray*}$ -dimensional rhombus" with half diagonal length (or how you call it...) $\begin{eqnarray*} m \end{eqnarray*}$ .
Then its perimeter/area/... grows polynomially (of degree $\begin{eqnarray*} n-1 \end{eqnarray*}$ ) with $\begin{eqnarray*} m \end{eqnarray*}$ .

Well, that's not a proof... But I actually don't know how to prove it properly Ups

But assuming that gives a majorant for
$\begin{eqnarray*} \sum_{k\in\mathbb Z^n}\frac C{(1+\lvert k\rvert)^{n+\varepsilon}}=\sum_{m=0}^\infty\frac{Cg(m)}{(1+m)^{n+\varepsilon}} \end{eqnarray*}$
of the form
$\begin{eqnarray*} \sum_{m=0}^\infty\frac{C\tilde Cm^{n-1}}{(1+m)^{n+\varepsilon}}\,. \end{eqnarray*}$

13.04.2013, 19:41

Lurchi_der_Lurch

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Zitat:

Original von Che Netzer

But assuming that gives a majorant for
$\begin{eqnarray*} \sum_{k\in\mathbb Z^n}\frac C{(1+\lvert k\rvert)^{n+\varepsilon}}=\sum_{m=0}^\infty\frac{Cg(m)}{(1+m)^{n+\varepsilon}} \end{eqnarray*}$
of the form
$\begin{eqnarray*} \sum_{m=0}^\infty\frac{C\tilde Cm^{n-1}}{(1+m)^{n+\varepsilon}}\,. \end{eqnarray*}$

to be honest i cannot see why $\begin{eqnarray*} g(m)\leq\tilde{C}m^{n-1} \end{eqnarray*}$ and it's a little bit unsatisfying

and now i guess the task is to show that

$\begin{eqnarray*} \sum_{m=0}^\infty\frac{C\tilde Cm^{n-1}}{(1+m)^{n+\varepsilon}}<\infty \end{eqnarray*}$

should i use the integral-comparison-criterion?

or can it be shown elementary?

13.04.2013, 20:46

Che Netzer

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Zitat:

Original von Lurchi_der_Lurch
to be honest i cannot see why $\begin{eqnarray*} g(m)\leq\tilde{C}m^{n-1} \end{eqnarray*}$ and it's a little bit unsatisfying

Sorry, I don't have enough time to think of a proof for that now.

Zitat:

should i use the integral-comparison-criterion?

or can it be shown elementary?

Yes, if you know that
$\begin{eqnarray*} \sum_{m=1}^\infty\frac1{m^\alpha} \end{eqnarray*}$
converges for $\begin{eqnarray*} \alpha>1 \end{eqnarray*}$ .

13.04.2013, 21:10

Slash123

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Zitat:

Original von Lurchi_der_Lurch
to be honest i cannot see why $\begin{eqnarray*} g(m)\leq\tilde{C}m^{n-1} \end{eqnarray*}$ and it's a little bit unsatisfying

Che gave me the permission to answer in this thread. I hope that there are no mistakes in my thoughts.

For $\begin{eqnarray*} m \in \mathbb{N} \end{eqnarray*}$ and $\begin{eqnarray*} n \end{eqnarray*}$ being fixed, define $\begin{eqnarray*} K_m := \{ k \in \mathbb{Z}^n | |k| = m, \, k \geq 0 \} \end{eqnarray*}$ , where $\begin{eqnarray*} k \geq 0 \end{eqnarray*}$ means that every $\begin{eqnarray*} k_i \geq 0 \end{eqnarray*}$ .

Now define a similar function to $\begin{eqnarray*} g \end{eqnarray*}$ :

$\begin{eqnarray*} h: \mathbb{N} \to \mathbb{N} \end{eqnarray*}$ given by $\begin{eqnarray*} m \mapsto |K_m| \end{eqnarray*}$ .

Obviously, the $\begin{eqnarray*} g(m) \end{eqnarray*}$ suffice the following inequation for every $\begin{eqnarray*} m: \end{eqnarray*}$

$\begin{eqnarray*} g(m) \leq 2^n h(m) \end{eqnarray*}$

Therefore, it suffices to show that $\begin{eqnarray*} h \in O(m^{n-1}) \end{eqnarray*}$ . Showing this isn't terribly hard. Just consider $\begin{eqnarray*} h(1) = n \end{eqnarray*}$ and try to calculate $\begin{eqnarray*} h(m) \end{eqnarray*}$ . For this, it could be useful to think about a possibility of getting all the elements in $\begin{eqnarray*} K_m \end{eqnarray*}$ by combining elements in $\begin{eqnarray*} K_1 \end{eqnarray*}$ .

Greetings,
Slash Wink

15.04.2013, 08:47

Che Netzer

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I just had another idea:
Let's assume $\begin{eqnarray*} \lvert k\rvert=\max_{i=1,\dotsc,n}\lvert k_1\rvert \end{eqnarray*}$ , i.e. $\begin{eqnarray*} \lvert\cdot\rvert \end{eqnarray*}$ be the supremum norm (again, by equivalence of norms, it doesn't matter).
Here we have $\begin{eqnarray*} (2m+1)^n \end{eqnarray*}$ integer vectors in $\begin{eqnarray*} \mathbb Z^n \end{eqnarray*}$ with norm $\begin{eqnarray*} m \end{eqnarray*}$ or less (the set of these vectors is a cube of side length $\begin{eqnarray*} m+1+m \end{eqnarray*}$ ).
So the number of vectors with norm $\begin{eqnarray*} m>0 \end{eqnarray*}$ is $\begin{eqnarray*} (2m+1)^n-(2(m-1)+1)^n=(2m+1)^n-(2m-1)^n \end{eqnarray*}$ , which is obviously a polynomial of degree $\begin{eqnarray*} n-1 \end{eqnarray*}$ or less.

If you were supposed to use another norm, there is a constant $\begin{eqnarray*} C \end{eqnarray*}$ such that that norm is greater or equal $\begin{eqnarray*} C \end{eqnarray*}$ times "our" norm.

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Konvergenz einer Reihe zeigen

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