Konvergenz einer Reihe zeigen

Neue Frage »

lurchi_der_lurch Auf diesen Beitrag antworten »
Konvergenz einer Reihe zeigen
Meine Frage:
Hallo, ich hab folgende Reihe mit :

.

Ich soll zeigen, dass diese Reihe konvergiert.

Meine Ideen:
Gibt's ein mehrdimensionales Integralvergleichskriterium?
Lurchi_der_Lurch Auf diesen Beitrag antworten »

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
 
 
Che Netzer Auf diesen Beitrag antworten »

By you probably mean .
If so, let and use that the number of with is polynomially increasing in with degree .
By that you may deduce the convergence from the convergence of some "one dimensional series".
Lurchi_der_Lurch Auf diesen Beitrag antworten »

thank you for answering

to be honest i do not know what is meant with .

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

, whereas



is that possible?
Che Netzer Auf diesen Beitrag antworten »

Yes, that could be possible, too.
Anyway, both norms are equivalent, so it doesn't matter Augenzwinkern
Lurchi_der_Lurch Auf diesen Beitrag antworten »

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

?

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

for example when it is imo



but i do not know what it is in common for n
Che Netzer Auf diesen Beitrag antworten »

Zitat:
Original von Lurchi_der_Lurch
do you mean

?

Almost. There is some factor (the number of with ) missing which increases in .
Lurchi_der_Lurch Auf diesen Beitrag antworten »

yes i thought so

but i do not know how to determine this factor and where to put it in the right sum
Che Netzer Auf diesen Beitrag antworten »

Actually, the exact factor doesn't matter.
Instead, we can estimate it by for almost all and some .
Lurchi_der_Lurch Auf diesen Beitrag antworten »

sorry, i do not see why the number of the k with increases in .
Lurchi_der_Lurch Auf diesen Beitrag antworten »

hey again!

i thought about it and came to the following

let be the number of those which fullfill

then f.a.



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


with regards

Lurchi
Che Netzer Auf diesen Beitrag antworten »

Let's just assume that for almost every and some Augenzwinkern
Maybe the following thought helps: Consider the set as some kind of "dotted -dimensional rhombus" with half diagonal length (or how you call it...) .
Then its perimeter/area/... grows polynomially (of degree ) with .

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

of the form
Lurchi_der_Lurch Auf diesen Beitrag antworten »

Zitat:
Original von Che Netzer

But assuming that gives a majorant for

of the form


to be honest i cannot see why and it's a little bit unsatisfying

and now i guess the task is to show that




should i use the integral-comparison-criterion?

or can it be shown elementary?
Che Netzer Auf diesen Beitrag antworten »

Zitat:
Original von Lurchi_der_Lurch
to be honest i cannot see why 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

converges for .
Slash123 Auf diesen Beitrag antworten »

Zitat:
Original von Lurchi_der_Lurch
to be honest i cannot see why 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 and being fixed, define , where means that every .

Now define a similar function to :

given by .

Obviously, the suffice the following inequation for every



Therefore, it suffices to show that . Showing this isn't terribly hard. Just consider and try to calculate . For this, it could be useful to think about a possibility of getting all the elements in by combining elements in .

Greetings,
Slash Wink
Che Netzer Auf diesen Beitrag antworten »

I just had another idea:
Let's assume , i.e. be the supremum norm (again, by equivalence of norms, it doesn't matter).
Here we have integer vectors in with norm or less (the set of these vectors is a cube of side length ).
So the number of vectors with norm is , which is obviously a polynomial of degree or less.

If you were supposed to use another norm, there is a constant such that that norm is greater or equal times "our" norm.
Neue Frage »
Antworten »



Verwandte Themen

Die Beliebtesten »
Die Größten »
Die Neuesten »