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Cauchy Condensation Proof

  1. Jan 2, 2008 #1
    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
    diverges if the series

    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.
    Last edited: Jan 2, 2008
  2. jcsd
  3. Jan 2, 2008 #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.
  4. Jan 2, 2008 #3
    changed, and what is baby rudin?
  5. Jan 2, 2008 #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.
  6. Jan 3, 2008 #5
    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.
  7. Jan 3, 2008 #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.
  8. Jan 4, 2008 #7
    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 writen 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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