Results 1 to 7 of 7

Math Help - the twisting and turning infinite sum of...

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    128

    the twisting and turning infinite sum of...

    \frac{2}{5} - \frac{3}{10} + \frac{4}{15} - \frac{5}{20} + \frac{6}{25} - \frac{7}{30} + ...

    so I think the series looks like:

    \sum_{n=1}^{\infty}{\frac{(-1)^{n-1}*(n+1)}{5n}}

    looks a little bit geometric, the problem is to show that it either converges or diverges, we have the powerful limit comparison test at our disposal but I am not sure what to compare it to, or if there is an alternate path to solution,
    I have tried the integral, basic comparison and nth term tests...
    any help kindly appriciated!

    thank you!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    the limit of the general term doesn't exist, so your series diverges.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    Posts
    128
    the limit of the general term, the limit of n goes to infinity of {\frac{(-1)^{n-1}*(n+1)}{5n}}
    I don't see how it does not exist? it looks like an indeterminate form to me, if you apply L'H we still end up with a (-1) to the infinity term in the numerator?? Should I take the log of the limit?

    I hope am not missing something obvious!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2009
    Posts
    128

    Continued Confusion

    So I have re-written the sum as

    \sum_{n=2}^{\infty}{\frac{(-1)^{n}*(n)}{5n-5}}

    But I still can't see that the limit doesn't exist, (I am aware that if it doesn't exist, then the series diverges.)

    Does L'Hopital's rule work here, (does the limit comparison test work for a lower sum limit of n=2?) because I don't see any other way of showing the limit is non-existant, and showing, therefore, that the sum diverges!

    Sorry if I am missing something obvious!

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

  5. #5
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by matt.qmar View Post
    the limit of the general term, the limit of n goes to infinity of {\frac{(-1)^{n-1}*(n+1)}{5n}}
    I don't see how it does not exist? it looks like an indeterminate form to me, if you apply L'H we still end up with a (-1) to the infinity term in the numerator?? Should I take the log of the limit?

    I hope am not missing something obvious!
    Let a_n=\frac{(-1)^{n-1}(n+1)}{5n} (just so I don't have to keep writing it).

    As n\to\infty, \frac{n+1}{5n}\to\frac{1}{5}\neq0, so the sum diverges. Conceptually, for large enough n, \sum_{n>N} a_n\approx \frac{1}{5}-\frac{1}{5}+\frac{1}{5}...

    You see why that diverges?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2009
    Posts
    128
    thank you so much! I am not sure how I missed the (n+1)/5n.

    Still not exactly sure how (-1)^infinity was taken care of, but I suppose it doesn't matter (always either +/- 1)

    thanks again!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2009
    Posts
    82
    Quote Originally Posted by matt.qmar View Post
    Still not exactly sure how (-1)^infinity was taken care of, but I suppose it doesn't matter (always either +/- 1)
    It wasn't taken care of. The terms oscillate, and they do not approach 0; as Krizalid said, the limit does not exist.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. A simple but mind twisting one
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: February 26th 2011, 12:12 PM
  2. Turning Points
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 15th 2009, 12:54 AM
  3. Proof: infinite set minus a fintie set is infinite
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: February 24th 2009, 11:54 PM
  4. Turning a mgf into a pmf
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: September 29th 2008, 09:10 PM
  5. Turning a P.D.E to an O.D.E
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 10th 2008, 08:14 AM

Search Tags


/mathhelpforum @mathhelpforum