Results 1 to 12 of 12

Math Help - summation problem

  1. #1
    Senior Member
    Joined
    Feb 2008
    Posts
    410

    summation problem

    Hi, all. I need to evaluate the following:

    \frac{1}{k}\sum_{n = 0}^{\infty}(\frac{k-1}{k})^nn

    where k∈ℕ and k≠1

    I don't really know how to do much else than test whether or not a series converges or diverges; I don't know how to determine what it converges to.

    Thanks, guys!
    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 hatsoff View Post
    Hi, all. I need to evaluate the following:

    \frac{1}{k}\sum_{n = 0}^{\infty}(\frac{k-1}{k})^nn

    where k∈ℕ and k≠1

    I don't really know how to do much else than test whether or not a series converges or diverges; I don't know how to determine what it converges to.

    Thanks, guys!
    I will give you a hint

    Rewrite \frac{k-1}{k} as 1-\frac{1}{k}

    Now consider that since k\in\mathbb{N} that 1-\frac{1}{k}<1


    Get it now?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Quote Originally Posted by Mathstud28 View Post
    I will give you a hint

    Rewrite \frac{k-1}{k} as 1-\frac{1}{k}

    Now consider that since k\in\mathbb{N} that 1-\frac{1}{k}<1


    Get it now?
    It resembles a geometric series in that respect, but then each term is multiplied by n, which is not a constant. I don't know how to handle that.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by hatsoff View Post
    It resembles a geometric series in that respect, but then each term is multiplied by n, which is not a constant. I don't know how to handle that.
    Think derivative/integration
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Quote Originally Posted by Mathstud28 View Post
    Think derivative/integration
    I need another hint, sir.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by hatsoff View Post
    Hi, all. I need to evaluate the following:

    \frac{1}{k}\sum_{n = 0}^{\infty}(\frac{k-1}{k})^nn

    where k∈ℕ and k≠1

    I don't really know how to do much else than test whether or not a series converges or diverges; I don't know how to determine what it converges to.

    Thanks, guys!
    If |x| < 1,
    G(x) = \sum_{n=0}^{\infty} x^n = \frac1{1-x}

    G'(x) = \sum_{n=0}^{\infty} nx^{n-1} = \frac1{(1-x)^2}

    xG'(x) = \sum_{n=0}^{\infty} nx^{n} = \frac{x}{(1-x)^2}

    Now let x = 1 - \frac1{k}, since k \in \mathbb{N}, we have \frac1{k} \leq 1 and 0 < 1 - \frac1{k} < 1. So we can apply the infinte Geometric Progression formula

    \sum_{n=0}^{\infty} n(1 - \frac1{k})^{n} = \frac{1 - \frac1{k}}{(\frac1{k})^2}

    \frac1{k} \sum_{n=0}^{\infty} n(1 - \frac1{k})^{n} = \frac{1 - \frac1{k}}{\frac1{k}} = k - 1
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by hatsoff View Post
    I need another hint, sir.
    Oh I am sorry! I completely missed your response. Luckily for you The oh so competent Isomorphism replied
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Quote Originally Posted by Mathstud28 View Post
    Think derivative/integration
    Hmm. Are you telling me that the following is true:

    \int[\sum_{n=r}^{\infty}a_n]dx=\sum_{n=r}^{\infty}\int[a_ndx]

    ?

    If so, then:

    \frac{1}{k}\sum_{n = 0}^{\infty}(\frac{k-1}{k})^nn

    \frac{1}{k}\sum_{n = 1}^{\infty}(\frac{k-1}{k})^nn

    \frac{k-1}{k^2}\sum_{n = 1}^{\infty}(\frac{k-1}{k})^{n-1}n

    (\frac{k-1}{k^2})\frac{d}{dn}\sum_{n = 1}^{\infty}\int[(\frac{k-1}{k})^{n-1}ndn]

    (\frac{k-1}{k^2})\frac{d}{dn}\sum_{n = 1}^{\infty}(\frac{k-1}{k})^{n}

    EDIT: I guess I'm late to the party.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by hatsoff View Post
    Hmm. Are you telling me that the following is true:

    \int[\sum_{n=r}^{\infty}a_n]dx=\sum_{n=r}^{\infty}\int[a_ndx]

    ?

    If so, then:

    \frac{1}{k}\sum_{n = 0}^{\infty}(\frac{k-1}{k})^nn

    \frac{1}{k}\sum_{n = 1}^{\infty}(\frac{k-1}{k})^nn

    \frac{k-1}{k^2}\sum_{n = 1}^{\infty}(\frac{k-1}{k})^{n-1}n

    (\frac{k-1}{k^2})\frac{d}{dn}\sum_{n = 1}^{\infty}\int[(\frac{k-1}<br /> <br />
{k})^{n-1}ndn]

    (\frac{k-1}{k^2})\frac{d}{dn}\sum_{n = 1}^{\infty}(\frac{k-1}{k})^{n}

    EDIT: I guess I'm late to the party.
    Yes you are...but for your sake yes..In a power series

    \int\sum_{n=0}^{\infty}a_n(x-c){n}dx=\sum_{n=0}^{\infty}\int{a_n(x-c)^{n}}dx

    The same applies for derivatives

    \frac{D[\sum_{n=0}^{\infty}a_n(x-c)^{n}]}{dx}=\sum_{n=0}^{\infty}\frac{D[a_n(x-c)^{n}]}{dx}

    to prove it expand...differentiate/integrate....contract into an explicit form power series and compare
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Mathstud28 View Post
    Yes you are...but for your sake yes..In a power series

    \int\sum_{n=0}^{\infty}a_n(x-c){n}dx=\sum_{n=0}^{\infty}\int{a_n(x-c)^{n}}dx

    The same applies for derivatives

    \frac{D[\sum_{n=0}^{\infty}a_n(x-c)^{n}]}{dx}=\sum_{n=0}^{\infty}\frac{D[a_n(x-c)^{n}]}{dx}

    to prove it expand...differentiate/integrate....contract into an explicit form power series and compare
    The following phrase should be added to this: "Provided the relevant series converge." If they don't then these rules go out the window.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by topsquark View Post
    The following phrase should be added to this: "Provided the relevant series converge." If they don't then these rules go out the window.

    -Dan
    Yes! Thank you! Sometimes I have the classic "I already know it so I assume others do to" syndrome...not a good thing to have
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Thanks, guys! This was a big help.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Summation Problem
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: December 29th 2010, 04:23 PM
  2. summation problem.
    Posted in the LaTeX Help Forum
    Replies: 3
    Last Post: October 4th 2010, 04:14 AM
  3. Summation problem
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 2nd 2009, 05:50 PM
  4. summation problem
    Posted in the Pre-Calculus Forum
    Replies: 22
    Last Post: November 8th 2009, 12:09 AM
  5. Summation problem
    Posted in the Algebra Forum
    Replies: 5
    Last Post: August 27th 2007, 12:21 PM

Search Tags


/mathhelpforum @mathhelpforum