Results 1 to 4 of 4

Math Help - Stuck with a series problem!

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    4

    Stuck with a series problem!

    The series is cosh(n)/n!.

    I attempted the ratio test with lim n→∞ of:
    \frac{\frac{\cosh(n+1)}{(n+1)!}}{\frac{\cosh(n)}{n  !}}=\frac{\cosh(n+1)}{cosh(n)}(n+1)=\frac{e^(n+1)+  e^(-n-1)}{e^(n)+e^(-n)}(n+1)

    and for me this means that L>1 if evaluated when lim n→∞, and so not convergent at all. However, the problem clearly states to determine whether the series converges absolutely or conditionally - hence why my conclusion obviously has to be wrong, and I am making some sort of mistake. I hope you guys can help me see what I'm doing wrong here.

    Thank you for your time!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    Shouldn't you have (n + 1) in the denominator?
    p.s. use e^{.......} for exponents with more than one character.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2011
    Posts
    73
    Thanks
    1
    Quote Originally Posted by endreoc View Post
    The series is cosh(n)/n!.

    I attempted the ratio test with lim n→∞ of:
    \frac{\frac{\cosh(n+1)}{(n+1)!}}{\frac{\cosh(n)}{n  !}}=\frac{\cosh(n+1)}{cosh(n)}(n+1)=\frac{e^(n+1)+  e^(-n-1)}{e^(n)+e^(-n)}(n+1)

    and for me this means that L>1 if evaluated when lim n→∞, and so not convergent at all. However, the problem clearly states to determine whether the series converges absolutely or conditionally - hence why my conclusion obviously has to be wrong, and I am making some sort of mistake. I hope you guys can help me see what I'm doing wrong here.

    Thank you for your time!
    All the terms for the cosh should be positive, no? I don't know why it would be asking you about absolute convergence.
    Last edited by eulcer; April 7th 2011 at 11:54 AM. Reason: Misread
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Quote Originally Posted by endreoc View Post
    The series is cosh(n)/n!.

    I attempted the ratio test with lim n→∞ of:
    \frac{\frac{\cosh(n+1)}{(n+1)!}}{\frac{\cosh(n)}{n  !}}=\frac{\cosh(n+1)}{cosh(n)}(n+1)=\frac{e^(n+1)+  e^(-n-1)}{e^(n)+e^(-n)}(n+1)

    and for me this means that L>1 if evaluated when lim n→∞, and so not convergent at all. However, the problem clearly states to determine whether the series converges absolutely or conditionally - hence why my conclusion obviously has to be wrong, and I am making some sort of mistake. I hope you guys can help me see what I'm doing wrong here.

    Thank you for your time!
    First step is to verify if \lim_{n \rightarrow \infty} \frac{\cosh n}{n!}=0 ... this scope is positively achieved remembering that is...

    \ln \cosh n \sim n - \ln 2

     \ln n! \sim \frac{\ln n}{2} + \frac{\ln 2 \pi}{2} + n \ln n - n (1)

    The second step [ratio test] is easy if You take into account that is...

    \cosh (n+1) = \cosh n\ \cosh 1 + \sinh n \sinh 1 \implies

    \implies \frac{\cosh (n+1)}{\cosh n} = \cosh 1 + \sinh 1 \tanh n (2)

    ... and of course that the term n+1 is in the denominator...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Getting stuck on series - can't develop Taylor series.
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: October 5th 2010, 08:32 AM
  2. Stuck on evaluating a series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 14th 2010, 09:24 PM
  3. I'm stuck on this series as well :(
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 13th 2010, 10:36 AM
  4. GP Series Problem...stuck
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: April 29th 2009, 09:22 AM
  5. Binomial series #2 stuck!
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 16th 2009, 10:09 PM

Search Tags


/mathhelpforum @mathhelpforum