Switching terms in a series - theorem

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  • #1
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Suppose a_n defined in the following way:
[tex]b_{2n}=a_{2n-1} [/tex]
[tex]b_{2n-1}=a_{2n} [/tex]
I know that [tex] \sum a_n [/tex] is convergent.

This how I proved that [tex] \sum b_n [/tex] is also convergent.

[tex]S_k=\sum_{n=1}^k b_n = \sum_{n=1}^k b_{2n} + \sum_{n=1}^k b_{2n-1} = \sum_{n=1}^k a_{2n-1} + \sum_{n=1}^k a_{2n} = \sum_{n=1}^k a_n \leq M [/tex]

Am I right?
 

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  • #2
HallsofIvy
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So you are saying that [itex]a_1+ a_2+ a_3+ a_4+ a_5+ a_6+ ...[/itex] becomes [itex]a_2+ a_1+ a_4+ a_3+ a_6+ a_5+ ...[/itex]?

What you have proved is that the sequence of partial sums is bounded above. If those partial sums are not increasing (if the [itex]a_n[/itex] are not all non-negative) that does not prove convergence.
 
  • #3
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Thanks, I now understand my mistake, but I can't find contra-example for this theorem.
My intuition locks me into thinking that switching 2 closest terms won't change convergence.
How should I approach such problems?
I will try thinking about Cauchy Criterion.
 
  • #4
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[tex] \sum a_n \rightarrow L\ \Rightarrow\ a_n\rightarrow0\ \Rightarrow\ \forall\ n>N_1\ a_n<\epsilon/4[/tex]
[tex] \sum a_n \rightarrow L\ \Rightarrow\ \forall\ m,n>N_2 |\\ \sum_{n+1}^m a_n|\leq \epsilon/4 [/tex]

So for all n>max{N_1+1,N_2+1}
[tex]|\sum_{k=n+1}^m b_k| \leq |a_n + \sum_{k=n+1}^m a_k + a_{m+1}| \leq \epsilon [/tex]

Is this idea right?
 
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  • #5
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I also try to explain in words my idea as I getting hard time with latex:

Because [tex]\sum a_n [/tex] is convergent and thanks to Cauchy Criterion we know that after some n I can sum a_n as many times as I want while keep the sum small as I want.
So I took little more terms to express [tex]\sum b_n[/tex] and to meet the Cauchy Criterion again.
I hope I expressed myself clearly
 
  • #6
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Still not sure about this, will appreciate opinions.
Thanks
 

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