Results 1 to 4 of 4

Math Help - cool series

  1. #1
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1

    cool series

    Mathstud likes series, as well as others. I enjoy a good one too. Here is one I thought was cool because of its familiar punchline.

    "For n=1,2,3,....... let c_{n}=1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{n}

    Evaluate the sum:

    S=\sum_{n=1}^{\infty}\frac{c_{n}}{n(n+1)}"

    Not that this is particularly difficult. I just liked it. Enjoy the holiday for those of you who celebrate the Fourth of July. Even if you're not American, take the time to enjoy anyway. Moo, you have Bastille Day coming up I think July 14th?. Do the French use it as an excuse to party as Americans do the fourth?.

    If no one bites by tomorrow morning, I will post my attempt.
    Last edited by galactus; July 3rd 2008 at 07:23 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by galactus View Post
    Mathstud likes series, as well as others. I enjoy a good one too. Here is one I thought was cool because of its familiar punchline.

    "For n=1,2,3,....... let c_{n}=1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{n}

    Evaluate the sum:

    S=\sum_{n=1}^{\infty}\frac{c_{n}}{n(n+1)}"

    Not that this is particularly difficult. I just liked it. Enjoy the holiday for those of you who celebrate the Fourth of July. Even if you're not American, take the time to enjoy anyway. Moo, you have Bastille Day coming up I think July 14th?. Do the French use it as an excuse to party as Americans do the fourth?.

    If no one bites by tomorrow morning, I will post my attempt.
    We can prove that \lim_{n \to \infty} \frac{c_{n}}{n} = 0.

    S=\sum_{n=1}^{\infty}\frac{c_{n}}{n(n+1)} = \sum_{n=1}^{\infty}\left(\dfrac{c_n}{n} - \frac{c_{n+1}- \frac{1}{n+1}}{n+1}\right) =\sum_{n=1}^{\infty}\left(\dfrac{c_n}{n} - \frac{c_{n+1}}{n+1}\right)+\sum_{n=1}^{\infty} \frac1{(n+1)^2}

    Clearly the first part telescopes to give:

    \sum_{n=1}^{\infty}\left(\dfrac{c_n}{n} - \frac{c_{n+1}}{n+1}\right) = c_1(=1) - \lim_{n \to \infty} \frac{c_{n+1}}{n+1}

    Now \lim_{n \to \infty} \frac{c_{n+1}}{n+1} = \lim_{n \to \infty} \frac{c_n}{n}=0

    Thus
    S = 1  + \sum_{n=1}^{\infty} \frac1{(n+1)^2} = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi ^2}{6}

    Note: Proof of \lim_{n \to \infty} \frac{c_n}{n}=0

    \lim_{n \to \infty} (c_n - \log n) = \gamma \Rightarrow \lim_{n \to \infty} \frac{c_n}{n} = \lim_{n \to \infty} \frac{\log n}{n} = 0
    Last edited by Isomorphism; July 3rd 2008 at 11:55 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Well done, Iso. Notice the familiar ending?. The ever popular Basel problem solution. Here is what I done. It's pretty much your approach.

    \sum_{n=1}^{k}\frac{c_{n}}{n(n+1)}

    =\sum_{n=1}^{k}\left(\frac{c_{n}}{n}-\frac{c_{n}}{n+1}\right)

    =\sum_{n=1}^{k}\frac{c_{n}}{n}-\sum_{n=2}^{k+1}\frac{c_{n-1}}{n}

    =c_{1}+\sum_{n=2}^{k}\frac{c_{n}-c_{n-1}}{n}-\frac{c_{k}}{k+1}

    =1+\sum_{n=2}^{k}\frac{1}{n^{2}}-\frac{c_{k}}{k+1}

    =\sum_{n=1}^{k}\frac{1}{n^{2}}-\frac{c_{k}}{k+1}

    =\sum_{n=1}^{k}\frac{1}{n^{2}}-\frac{c_{k}-ln(k)}{k+1}-\frac{ln(k)}{k+1}

    Letting n\to {\infty}, and using the fact that:

    \lim_{k\to {\infty}}(c_{k}-ln(k)) exists and

    \lim_{k\to {\infty}}\frac{ln(k)}{k+1}=0

    Finally:

    S=\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\boxed{\frac{  {\pi}^{2}}{6}}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by galactus View Post
    Well done, Iso. Notice the familiar ending?. The ever popular Basel problem solution.
    Of course, how else would I have computed that final sum

    Actually I was trying a different approach too. I worked it out and posted it in the Definitely Integral thread. Actually your sum amounts to this integral:
    \int_0^1 \int_0^x \frac{\ln (1 - y)}{y(y-1)}   ~ dy ~ dx and its easy to work this out. Check out wingless's solution.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cool forum! I need a little help....
    Posted in the Statistics Forum
    Replies: 2
    Last Post: February 20th 2010, 02:57 PM
  2. Very cool
    Posted in the Calculus Forum
    Replies: 11
    Last Post: December 21st 2009, 11:32 AM
  3. A few cool ones here
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: March 26th 2009, 06:00 PM
  4. cool integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: December 29th 2008, 01:07 PM
  5. cool theorem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 14th 2008, 04:50 PM

Search Tags


/mathhelpforum @mathhelpforum