Cauchy Condensation Proof

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Discussion Overview

The discussion revolves around the Cauchy condensation test for series, specifically examining the conditions under which the series \(\sum^{\infty}_{n=1}b_{n}\) diverges given that the series \(\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}\) diverges. Participants explore various approaches and reasoning related to this mathematical proof.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes to prove the divergence of \(\sum^{\infty}_{n=1}b_{n}\) by contradiction, assuming it converges and showing that \(\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}\) must diverge.
  • Another participant suggests that the series \(\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}\) converges if \(\sum^{\infty}_{n=1}b_{n}\) converges, indicating a contrapositive relationship.
  • A different viewpoint emphasizes rearranging the series to demonstrate that the condensed series is no worse than a multiple of the original series.
  • Several participants discuss the implications of the series being decreasing and non-negative, with one referencing a proof from "baby Rudin" as a source of established knowledge.
  • One participant elaborates on their reasoning involving partial sums and inequalities, attempting to connect the divergence of the condensed series to the divergence of the original series.

Areas of Agreement / Disagreement

Participants express differing opinions on the approach to proving the divergence of the series. There is no consensus on the best method or the correctness of the reasoning presented, indicating that multiple competing views remain.

Contextual Notes

Some participants note the importance of the conditions under which the series are defined, such as the non-negativity and monotonicity of the sequence \((b_{n})\). There are also unresolved mathematical steps in the reasoning provided by participants.

linuxux
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Hello. This is the presented problem:

Suppose [tex](b_{n})[/tex] is a decreasing satisfying [tex]b_{n}\ge\ 0[/tex]. Show that the series
[tex]\sum^{\infty}_{n=1}b_{n}[/tex]
diverges if the series
[tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}[/tex]
diverges.

I've already proved that i can create [tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}[/tex] from [tex]\sum^{\infty}_{n=1}b_{n}[/tex] and that that series is larger, so my first idea is to prove this by finding some kind of contradiction by supposing [tex]\sum^{\infty}_{n=1}b_{n}[/tex] converges and trying to prove that the series [tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}[/tex] diverges.

my idea is if [tex]\sum^{\infty}_{n=1}b_{n}[/tex] is bounded by [tex]M[/tex], then the worst case scenario for [tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}\ is\ b_{1}\ +\ ...\ +\ 2^{n}M[/tex] but we know there must be something greater since it is unbounded.

i'm not sure if that reasoning works or not.

thanks for the help.
 
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I think it must be b_n >= 0. Just letting you know that this is proved in baby Rudin, chapter 3: sequences and series.
 
rudinreader said:
I think it must be b_n >= 0. Just letting you know that this is proved in baby Rudin, chapter 3: sequences and series.

changed, and what is baby rudin?
 
Principles of Mathematical Analysis, 3rd Ed. Walter Rudin. It's a pervasive but nonetheless not uniformly liked book, so it's always in the library.
 
linuxux said:
my first idea is to prove this by finding some kind of contradiction by supposing [tex]\sum^{\infty}_{n=1}b_{n}[/tex] converges and trying to prove that the series [tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}[/tex] diverges.

You mean to show that [tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}[/tex] converges correct? This is just the contrapositive of the statement.

Just a hunch. I'm not too well versed with sequences and series, but it looks to me like some sort of root test would be due here.
 
No, no, you want to rearrange the series in a different way to get that the condensed series is no worse than a multiple of the original.
 
linuxux said:
Hello. This is the presented problem:

Suppose [tex](b_{n})[/tex] is a decreasing satisfying [tex]b_{n}\ge\ 0[/tex]. Show that the series
[tex]\sum^{\infty}_{n=1}b_{n}[/tex]
diverges if the series
[tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}[/tex]
diverges.

I've already proved that i can create [tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}[/tex] from [tex]\sum^{\infty}_{n=1}b_{n}[/tex] and that that series is larger, so my first idea is to prove this by finding some kind of contradiction by supposing [tex]\sum^{\infty}_{n=1}b_{n}[/tex] converges and trying to prove that the series [tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}[/tex] diverges.

my idea is if [tex]\sum^{\infty}_{n=1}b_{n}[/tex] is bounded by [tex]M[/tex], then the worst case scenario for [tex]\sum^{\infty}_{n=0}{2^{n}b_{2^{n}}}\ is\ b_{1}\ +\ ...\ +\ 2^{n}M[/tex] but we know there must be something greater since it is unbounded.

i'm not sure if that reasoning works or not.

thanks for the help.

since [tex]\{b_n}[/tex] is a decreasing series we have b1>b2>b3>b4>...>bn
[tex]\sum^{\infty}_{k=0}{2^{k}b_{2^{k}}}[/tex] now let this series diverge,

first let us suppose that n>2^k, now let us take the partial sums of both series

let us denote by Sn the partial sum of the series [tex]\sum^{\infty}_{n=1}b_{n}[/tex]
we have Sn=b1+b2+b3+b4+...+bn>1/2b1+b2+(b3+b4)+(b5+b6+b7+b8)+...+(b_2^(k-1)+1...+b_2^k)>1/2 b1 +b2+2b4+4b8+...+2^(k-1)b_2^k), now we multiply both sides by 2 and we get on the right side

b1+2b2+4b4+8b8+...+2^kb_2^k , let us denote this by W_k, so

W_k=b1+2b2+4b4+8b8+...+2^kb_2^k, which actually is the partial sum of the series
[tex]\sum^{\infty}_{k=0}{2^{k}b_{2^{k}}}[/tex], now eventually we have

2Sn>W_k, so assuming that [tex]\sum^{\infty}_{k=0}{2^{k}b_{2^{k}}}[/tex], diverges we know that its partial sum is not upper bounded, moreover if we take the limit of its partial sum it converges to infinity, so it automatically yelds that also the partial sum Sn of the series [tex]\sum^{\infty}_{n=1}b_{n}[/tex], is not upper bounded, moreover the limit of it as n-->infinity the partial sum also converges to infinity, so the series [tex]\sum^{\infty}_{n=1}b_{n}[/tex] also diverges.

i appologize for my symbols, and i do not know whether i have written all the neccesary elements, because i did it very quickly now, but anyway this is the idea of prooving this.

i hope i was of any help
 

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