Konvergenz einer Reihe zeigen 
13.04.2013, 13:57  lurchi_der_lurch  Auf diesen Beitrag antworten »  
Konvergenz einer Reihe zeigen Hallo, ich hab folgende Reihe mit : . Ich soll zeigen, dass diese Reihe konvergiert. Meine Ideen: Gibt's ein mehrdimensionales Integralvergleichskriterium? 

13.04.2013, 15:26  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 

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13.04.2013, 15:40  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". 

13.04.2013, 15:45  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? 

13.04.2013, 15:51  Che Netzer  Auf diesen Beitrag antworten »  
Yes, that could be possible, too. Anyway, both norms are equivalent, so it doesn't matter 

13.04.2013, 15:55  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. 

13.04.2013, 16:33  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 

13.04.2013, 17:03  Che Netzer  Auf diesen Beitrag antworten »  
Almost. There is some factor (the number of with ) missing which increases in . 

13.04.2013, 17:05  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 

13.04.2013, 17:09  Che Netzer  Auf diesen Beitrag antworten »  
Actually, the exact factor doesn't matter. Instead, we can estimate it by for almost all and some . 

13.04.2013, 17:23  Lurchi_der_Lurch  Auf diesen Beitrag antworten »  
sorry, i do not see why the number of the k with increases in . 

13.04.2013, 19:19  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 

13.04.2013, 19:34  Che Netzer  Auf diesen Beitrag antworten »  
Let's just assume that for almost every and some 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 But assuming that gives a majorant for of the form 

13.04.2013, 19:41  Lurchi_der_Lurch  Auf diesen Beitrag antworten »  
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 integralcomparisoncriterion? or can it be shown elementary? 

13.04.2013, 20:46  Che Netzer  Auf diesen Beitrag antworten »  
Sorry, I don't have enough time to think of a proof for that now.
Yes, if you know that converges for . 

13.04.2013, 21:10  Slash123  Auf diesen Beitrag antworten »  
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 

15.04.2013, 08:47  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. 

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