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convergent sequances
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lomos_lupin
Yang-Mills Theory
Yang-Mills Theory


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#1
convergent sequances
just a question

What are the suffice and necessary conditions for a sequance like \{a_i\} such that

1.\sum_{i=1}^{\infty} x_i= convergent.

2.\sum_{i=1}^{\infty}(-1)^i x_i= convergent



2. ,my teacher told me : these 2 conditions are enought but i didnt managed to proof.So would someone confirme the rightness of this?
1.lim|a_i| \longrightarrow 0 when i \longrightarrow \infty
2.|a_{i+1} | \le |a_i|
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Last edited by lomos_lupin on Sun Sep 11, 2005 5:35 am; edited 2 times in total 
PostPosted: Sun Sep 11, 2005 4:06 am  Back to top 
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Diarmuid
Poincare Conjecture
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#2
The answer you have for (1) is wrong - the classical counterexample being a_i=\frac{1}{i}, which is decreasing and limits to zero as in the conditions you mention, but the sum of whose terms diverges.

PostPosted: Sun Sep 11, 2005 5:29 am  Back to top 
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lomos_lupin
Yang-Mills Theory
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#3
Diarmuid wrote:
The answer you have for (1) is wrong - the classical counterexample being a_i=\frac{1}{i}, which is decreasing and limits to zero as in the conditions you mention, but the sum of whose terms diverges.

I ment this for \sum_{i=1}^{\infty} (-1)^i x_i ,i have edited the post .

By the way ,For a_i=\frac1i what will be \sum_{i=1} (-1)^i a_i Question
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PostPosted: Sun Sep 11, 2005 5:38 am  Back to top 
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Cezar Lupu
Navier-Stokes Equations
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#4
it will be ln 2 Wink
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PostPosted: Sun Sep 11, 2005 5:47 am  Back to top 
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Diarmuid
Poincare Conjecture
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#5
Well it's not really right for the second one either - e.g. a_i=\frac{(-1)^i}{i} gives \sum_{i=1}^\infty(-1)^ia_i=\sum_{i=1}^\infty\frac{1}{i}, which is the same thing.

If you additionally assume that all the terms are positive (or negative), then you'd have a sufficient condition.

PostPosted: Mon Sep 12, 2005 1:16 am  Back to top 
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blahblahblah
Birch & Swinnerton Dyer
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#6
1) The only reasonable sufficient condition you're going to get is something like the Cauchy criterion, unless you impose much tighter restrictions on \{a_i\}. In particular, a_i\to 0 is not strong enough.

2) If each a_i is positive, then it is enough to have a_i\geq a_j for i\leq j, and a_i\to 0. This is the alternating series test; you may prove it using partial summation.

PostPosted: Mon Sep 12, 2005 1:29 am  Back to top 
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perfect_radio
Birch & Swinnerton Dyer
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#7
i have a related question: knowing that \sum_{i=1}^{\infty} |a_i| converges, how do you prove that \sum_{i=1}^{\infty} a_i converges?
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PostPosted: Fri Sep 16, 2005 10:17 am  Back to top 
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Diarmuid
Poincare Conjecture
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#8
Triangle inequality on the partial sums?

(i.e. \left|\sum a_i\right|\le\sum|a_i|).

PostPosted: Fri Sep 16, 2005 11:32 am  Back to top 
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perfect_radio
Birch & Swinnerton Dyer
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#9
Diarmuid wrote:
Triangle inequality on the partial sums?

(i.e. \left|\sum a_i\right|\le\sum|a_i|).


is this inequality sufficient to prove it? i didn't manage to prove it from here Sad Blush
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PostPosted: Sun Oct 02, 2005 1:54 pm  Back to top 
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blahblahblah
Birch & Swinnerton Dyer
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#10
Yes, it is enough. Try using the Cauchy criterion.

PostPosted: Sun Oct 02, 2005 2:13 pm  Back to top 
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