Results 1 to 5 of 5

Math Help - Convergence/Divergence of Series

  1. #1
    Member
    Joined
    Nov 2006
    From
    San Francisco
    Posts
    145

    Convergence/Divergence of Series

    Can someone give me a hint that the summation of (-1)^(u)*(u^(1/2))/(u+1) from u=1 to N is divergent, conditionally convergent, or absolutely convergent? By the looks of it it is conditionally convergent as it's absolute value diverges, but i know 1/u^(1/2)*(-1)^(u) converges and the one doesn't really matter, but I can't prove that it doesn't matter.

    Thanks for the help in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    42
    Quote Originally Posted by jarny View Post
    Can someone give me a hint that the summation of (-1)^(u)*(u^(1/2))/(u+1) from u=1 to N is divergent, conditionally convergent, or absolutely convergent? By the looks of it it is conditionally convergent as it's absolute value diverges, but i know 1/u^(1/2)*(-1)^(u) converges and the one doesn't really matter, but I can't prove that it doesn't matter.

    Thanks for the help in advance
    You're right. It's conditionally convergent.

    Your sum is \sum_{i=1}^\infty \frac{(-1)^n \sqrt{n}}{n+1}. Considering \sum_{i=1}^\infty \frac{\sqrt{n}}{n+1} show it's divergent by LCT with \sum_{i=1}^\infty \frac{1}{\sqrt{n}}.

    The Alternating series test

    (i) \lim_{n \to \infty} \frac{\sqrt{n}}{n+1} = 0
    (ii) \frac{\sqrt{n}}{n+1} is decreasing

    give that it converges and thus, converges conditionally.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,739
    Thanks
    645
    Hello, jarny!

    You've already worked out the answer . . .


    \sum^{\infty}_{u=1}\;(\text{-}1)^u\frac{u^{\frac{1}{2}}}{u+1} . Divergent, conditionally convergent, or absolutely convergent?
    It is a convergent alternating series. .(The n^{th} term approaches zero.)

    It is not absolutely convergent because: . \sum\frac{u^{\frac{1}{2}}} {u+1} \:\approx\:\sum\frac{1}{u^{\frac{1}{2}}} . . . a divergent p-series.

    Therefore, the series is conditionally convergent.

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2006
    From
    San Francisco
    Posts
    145
    Somehow I managed to mis-write it. It in the denominator is should be 4 instead of 1. Sorry :/ Soroban, I can see that it is approximately but I need to kind of prove it.


    when its (-1)^(u) *(u)^(1/2)/(u+4) could I just find a particular u (in this case 4) where the absolute value of the series begins to decrease and say since a_1>a_2>a_3 ...>a_N and since the lim n->infinity is zero (and the summands before 4 are finite) that by the alternating series test the summation is conditionally convergent?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    42
    Quote Originally Posted by jarny View Post
    Somehow I managed to mis-write it. It in the denominator is should be 4 instead of 1. Sorry :/ Soroban, I can see that it is approximately but I need to kind of prove it.
    If you mean

    <br />
\sum_{i=1}^\infty \frac{(-1)^n \sqrt{n}}{n+4}

    It won't change the results that we have posted (except for the 4)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. series convergence / divergence
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 11th 2011, 09:08 PM
  2. [SOLVED] Divergence or Convergence of Series
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: January 26th 2011, 10:54 AM
  3. series convergence and divergence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 30th 2008, 03:19 PM
  4. Convergence/Divergence of Series.
    Posted in the Calculus Forum
    Replies: 12
    Last Post: November 14th 2008, 06:33 PM
  5. Convergence/Divergence of a series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 4th 2008, 05:41 PM

Search Tags


/mathhelpforum @mathhelpforum