Hello. This is the presented problem:(adsbygoogle = window.adsbygoogle || []).push({});

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.

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Cauchy Condensation Proof

Loading...

Similar Threads - Cauchy Condensation Proof | Date |
---|---|

I Cauchy Principal Value | Mar 23, 2017 |

I Cauchy Reimann Condition | Aug 22, 2016 |

I Cauchy's Integral Test for Convergence | Jun 20, 2016 |

I Cauchy Repeated Integration Explanation? | Mar 11, 2016 |

Question on the Cauchy Condensation test | Feb 6, 2008 |

**Physics Forums - The Fusion of Science and Community**