Cauchy Condensation Proof

In summary, if the series \sum^{\infty}_{n=0}{2^{n}b_{2^{n}}} diverges, then the series \sum^{\infty}_{n=1}b_{n} also diverges.
  • #1
linuxux
133
0
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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  • #2
I think it must be b_n >= 0. Just letting you know that this is proved in baby Rudin, chapter 3: sequences and series.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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
 

1. What is Cauchy Condensation Proof?

Cauchy Condensation Proof is a mathematical proof technique used to prove the convergence or divergence of infinite series. It was developed by the French mathematician Augustin-Louis Cauchy in the 19th century.

2. How does Cauchy Condensation Proof work?

This proof technique involves comparing the convergence of a given series with a new series formed by condensing the terms of the original series. If the condensed series converges, then the original series also converges. If the condensed series diverges, then the original series also diverges.

3. What is the advantage of using Cauchy Condensation Proof?

The main advantage of using Cauchy Condensation Proof is that it simplifies the process of determining the convergence or divergence of an infinite series. It also allows for the comparison of series that may not have the same terms, making it a versatile tool in mathematical analysis.

4. What are the limitations of Cauchy Condensation Proof?

One limitation of this proof technique is that it only works for series with positive terms. It also cannot be used to prove absolute convergence. Additionally, it may not always provide a definitive answer, as some series may have convergence behavior that is not reflected in the condensed series.

5. In what areas of mathematics is Cauchy Condensation Proof commonly used?

Cauchy Condensation Proof is commonly used in real analysis, complex analysis, and numerical analysis. It is also frequently used in calculus and in the study of infinite series in general.

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