Results 1 to 3 of 3

Math Help - Convergence of a series.

  1. #1
    s3a
    s3a is offline
    Super Member
    Joined
    Nov 2008
    Posts
    597

    Convergence of a series.

    I am supposed to show that it converges or diverges. If it diverges, I need to further demonstrate whether it converges absolutely or conditionally.

    Here is the problem I am having trouble with so far that I happened to get right (through the use of my logic but I don't know how to write it in math terms - I'll explain in detail as you read):

    Problem: sum of ((n^2 + 1)/(2n^2 + 1))^n from 1 to inf - Wolfram|Alpha

    My logic is that the limit as n->inf of (n^2 + 1)/(2n^2 + 1) is 1/2 and (1/2)^n is a geometric series and it therefore should converge but this seems to be not good enough to me because of the fact that the first term for isntance is 2/3 and the second one is 5/9 which are somewhat away from 1/2 but then on the other hand the part of my logic that says it is still okay is that even the first term even though it's not near 1/2 is less than 1 which, in a geometric series, implies convergence.

    Any input would be GREATLY appreciated!
    Thanks in advance!
    Last edited by mr fantastic; May 6th 2010 at 06:20 PM. Reason: Added url tags to link, re-titled.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member apcalculus's Avatar
    Joined
    Apr 2009
    From
    Boston
    Posts
    293
    Quote Originally Posted by s3a View Post
    I am supposed to show that it converges or diverges. If it diverges, I need to further demonstrate whether it converges absolutely or conditionally.

    Here is the problem I am having trouble with so far that I happened to get right (through the use of my logic but I don't know how to write it in math terms - I'll explain in detail as you read):

    Problem: sum of ((n^2 + 1)/(2n^2 + 1))^n from 1 to inf - Wolfram|Alpha

    My logic is that the limit as n->inf of (n^2 + 1)/(2n^2 + 1) is 1/2 and (1/2)^n is a geometric series and it therefore should converge but this seems to be not good enough to me because of the fact that the first term for isntance is 2/3 and the second one is 5/9 which are somewhat away from 1/2 but then on the other hand the part of my logic that says it is still okay is that even the first term even though it's not near 1/2 is less than 1 which, in a geometric series, implies convergence.

    Any input would be GREATLY appreciated!
    Thanks in advance!
    Because the typical term in this series is an expression raised to the power n, you should use the root test.

    \lim_{n\to\infty}\sqrt[n]{a_n} = L

    If L < 1, the series converges.
    If L > 1, the series diverges.
    If L = 1, the test is inconclusive.

    A summary of convergence/divergence test can be download here:
    http://abacus.bates.edu/acad/acad_su...ence_tests.pdf

    Good luck!!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    Quote Originally Posted by s3a View Post
    I am supposed to show that it converges or diverges. If it diverges, I need to further demonstrate whether it converges absolutely or conditionally.

    Here is the problem I am having trouble with so far that I happened to get right (through the use of my logic but I don't know how to write it in math terms - I'll explain in detail as you read):

    Problem: sum of &#x28;&#x28;n&#x5e;2 &#x2b; 1&#x29;&#x2f;&#x28;2n&#x5e;2 &#x2b; 1&#x29;&#x29;&#x5e;n from 1 to inf - Wolfram|Alpha

    My logic is that the limit as n->inf of (n^2 + 1)/(2n^2 + 1) is 1/2 and (1/2)^n is a geometric series and it therefore should converge but this seems to be not good enough to me because of the fact that the first term for isntance is 2/3 and the second one is 5/9 which are somewhat away from 1/2 but then on the other hand the part of my logic that says it is still okay is that even the first term even though it's not near 1/2 is less than 1 which, in a geometric series, implies convergence.

    Any input would be GREATLY appreciated!
    Thanks in advance!
    As Wolfram's Alpha tells you, one would ordinarily use the root test to figure it out, but your idea isn't bad either. Since, as you write q_n := \frac{n^2 + 1}{2n^2 + 1}\rightarrow \frac{1}{2}, for any \varepsilon > 0, no matter how small, you can find an n_0 such that for all n>n_0 your q_n is held in a "sandwich" 0\leq \left(\frac{1}{2}-\varepsilon\right)^n\leq q_n^n\leq \left(\frac{1}{2}+\varepsilon\right)^n.
    This implies convergence of the partial sums of your series, and thus of the series itself.

    P.S: In fact, the argument is even simpler than that: since the terms of your series are all positive, it suffices to show that the partial sums are bounded from above, thus is suffices that there exists an n_0 such that for all n>n_0 we have that 0\leq q_n^n \leq \left(\frac{3}{4}\right)^n.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: October 3rd 2011, 01:12 AM
  2. Replies: 2
    Last Post: May 1st 2010, 09:22 PM
  3. Replies: 4
    Last Post: December 1st 2009, 03:23 PM
  4. Replies: 3
    Last Post: December 1st 2009, 03:06 PM
  5. series convergence and radius of convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 15th 2008, 08:07 AM

Search Tags


/mathhelpforum @mathhelpforum