Results 1 to 2 of 2

Math Help - Complicated Series

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    17

    Complicated Series

    Consider the Series: ∑ 1/[2^(k+(-1)^k)] from k=0 to ∞.

    a. Use the ratio test to show the limit does not exist.

    b. Use the root test to show the series converges.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by jules027 View Post
    Consider the Series: ∑ 1/[2^(k+(-1)^k)] from k=0 to ∞.

    a. Use the ratio test to show the limit does not exist.

    b. Use the root test to show the series converges.
    \sum_{n=0}^{\infty}\frac{1}{2^{n+(-1)^n}}

    Setting up the ratio test we get

    \lim_{n\to\infty}\bigg|\frac{1}{2^{(n+1)(-1)^{n+1}}}\cdot\frac{2^{n+(-1)^n}}{1}\bigg|=\lim_{n\to\infty}|2^{-((n+1)+(-1)^{n+1})}\cdot{2^{n+(-1)^n}|}=\lim_{n\to\infty}2^{2(-1)^n-1}= \lim_{n\to\infty}\frac{4^{(-1)^n}}{2}


    Which is an oscilating function...to realize this also notice that (-1)^n=\cos(\pi{n})

    In this case only because n\in\mathbb{N}

    Next if we do the root test we get

    \lim_{n\to\infty}\bigg(2^{-(n+(-1)^n)}\bigg)^{\frac{1}{n}}=\lim_{n\to\infty}2^{-1-\frac{(-1)^n}{n}}=\lim_{n\to\infty}\frac{1}{2}\cdot{2^{\fr  ac{(-1)^{n+1}}{n}}}=\frac{1}{2}\cdot{2^{0}}=\frac{1}{2}  <1

    Thus since this is less than one the series converges
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complicated Function
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 1st 2010, 04:00 AM
  2. Complicated Geometric Series Question.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 10th 2009, 07:02 AM
  3. complicated p-series
    Posted in the Calculus Forum
    Replies: 9
    Last Post: March 22nd 2009, 05:04 PM
  4. Complicated Equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 26th 2009, 07:28 PM
  5. Complicated Log problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: December 18th 2008, 10:44 AM

Search Tags


/mathhelpforum @mathhelpforum