PuppiesOnAcid | Or would that even be possible Because as I understand it, the first term of your geometric series has to be just your "a" value |

kelt | PuppiesOnAcid: there is a rule (geometric Ithink) like for all p > 2 (1/p)^x converges? |

PuppiesOnAcid | That's not rly helping me here though |

Kasadkad | PuppiesOnAcid: sum(x^n, n=0..infinity) = 1/(1-x) if |x| < 1, that's the geometric series formula |

PuppiesOnAcid | Kasadkad: I have a geometric series as being sum(ar^(n-1), n=1..oo) |

Kasadkad | That's the same as a*sum(r^n, n=0..infinity) Just shift the indices You can pull the a out of the sum too; it doesn't affect convergence at all |

PuppiesOnAcid | Kasadkad: The summation I'm working with is sum(1/3^n, n=1..oo) Is that in geometric form? If so, what is the a and the r value? |

Kasadkad | r = 1/e er r = 1/3 a = 1 |

kelt | well if you do sqrt(b-4ac) = sqrt(1+4) = sqrt(5) so x^2+x-1 shouldn't have a complex number but x^2+x+1 should I guess |

PuppiesOnAcid | Right, but the first term of your geometric series has to be your a value |

Goplat | Kasadkad: actually since you start from n=1, a isn't 1 |

pengrate | kelt: Yeah, you can find the factors by using the quadratic formula |

Kasadkad | Ehm, yeah |

PuppiesOnAcid | Goplat: Exactly, that is my hang up |

Kasadkad | a = 1/3 then But it doesn't matter |

PuppiesOnAcid | if a = 1/3, is your r value still 1/3? |

Kasadkad | You don't need the exact form anyway |

kelt | pengrate: you solve for x? then do (x+a1)(x-a2) or whatever? |

PuppiesOnAcid | Kasadkad: I need the a and r values so I can find the sum |

pengrate | kelt: Well, (x-a1)(x-a2), but yeah |

Kasadkad | The geometric series formula tells you that sum(1/3^n, n=0..infinity) = 1/(1-1/3) = 3/2 But what's sum(1/3^n, n=1..infinity) in terms of that? |

Echelon | hmm if i say m E integers do i say m is either even or odd or |

su-hoens | hej |

Echelon | m is either even, off, or 0 |

su-hoens | hej |