Results 1 to 6 of 6

Math Help - Divergence or Convergence of Series

  1. #1
    Junior Member
    Joined
    Mar 2008
    From
    Sweden
    Posts
    61

    Divergence or Convergence of Series

    Hi,

    I have have to determine if the following series is convergent or divergent.

    \displaystyle \sum_{k= 1}^{\infty} \frac{(k!)^{\frac{1}{k}}}{k}}= \sum_{k= 1}^{\infty} \root k \of {k!}{\frac{1}{k}

    \displaystyle a_k=\frac{(k!)^{\frac{1}{k}}}{k}}

    \displaystyle b_k=\frac{1}{k}}}

    b_k is a p-series and diverges.

    \displaystyle\lim_{k \to \infty} \frac{a_k}{b_k}}=\root k \of {k!}=1

    According to direct comparison rule, since b_k is divergent then the given series must also be divergent.(1)


    Is (1)correct?

    I am also unsure about the following limit:

    (2) \displaystyle\lim_{k \to \infty} \sqrt{(k!)}=1


    Is (2) correct?

    I would appreciate a response.

    Thank you
    Last edited by CaptainBlack; January 26th 2011 at 05:34 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by 4Math View Post
    Hi,

    I have have to determine if the following series is convergent or divergent.

    \displaystyle \sum_{k= 1}^{\infty} \frac{(k!)^{\frac{1}{k}}}{k}}= \sum_{k= 1}^{\infty} \root k \of {k!}{\frac{1}{k}

    \displaystyle a_k=\frac{(k!)^{\frac{1}{k}}}{k}}

    \displaystyle b_k=\frac{1}{k}}}

    b_k is a p-series and diverges.

    \displaystyle\lim_{k \to \infty} \frac{a_k}{b_k}}=\root k \of {k!}=1

    According to direct comparison rule, since b_k is divergent then the given series must also be divergent.(1)


    Is (1)correct?

    I am also unsure about the following limit:

    (2) \displaystyle\lim_{k \to \infty} \sqrt{(k!)}=1


    Is (2) correct?

    I would appreciate a response.

    Thank you
    Why do you think that:

    \displaystyle  \lim_{n\to \infty} (n!)^{1/n}=1

    What light does Stirlings formula throw on this limit?

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2008
    From
    Sweden
    Posts
    61
    Quote Originally Posted by CaptainBlack View Post
    Why do you think that:

    \displaystyle  \lim_{n\to \infty} (n!)^{1/n}=1

    What light does Stirlings formula throw on this limit?

    CB

    The correct limit should be:

    \displaystyle\lim_{k \to \infty} \sqrt{(k!)}\rightarrow \infty when  k \to \infty

    Unfortunately I cannot see how Stilrlings formula can be applied and/or relate to this limit. Can you please show this.

    Kind regards
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Quote Originally Posted by 4Math View Post
    Unfortunately I cannot see how Stilrlings formula can be applied and/or relate to this limit. Can you please show this.

    Using Stirling's formula:


    \displaystyle\lim_{k \to{+}\infty}{\dfrac{\sqrt[k]{k!}}{k}}=\displaystyle\lim_{k \to{+}\infty}\dfrac{(2\pi k)^{\frac{1}{2k}}ke^{-1}}{k}=\ldots=\dfrac{1}{e}\neq 0

    So, the series is divergent.



    Fernando Revilla
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2008
    From
    Sweden
    Posts
    61
    Quote Originally Posted by FernandoRevilla View Post
    Using Stirling's formula:


    \displaystyle\lim_{k \to{+}\infty}{\dfrac{\sqrt[k]{k!}}{k}}=\displaystyle\lim_{k \to{+}\infty}\dfrac{(2\pi k)^{\frac{1}{2k}}ke^{-1}}{k}=\ldots=\dfrac{1}{e}\neq 0

    So, the series is divergent.



    Fernando Revilla
    Thank you for your response.

    Can I also use Direct Comparison Test like I have on my first post and if so have I done it correctly?

    Kind regards
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Quote Originally Posted by 4Math View Post
    Can I also use Direct Comparison Test like I have on my first post and if so have I done it correctly?
    Yes, you can. But,

    \displaystyle\lim_{k \to{+}\infty}{\dfrac{a_n}{b_n}}=+\infty

    and

    \displaystyle\sum_{k=1}^{+\infty}b_n

    is divergent, so the given series is divergent.


    Fernando Revilla
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] convergence / divergence of a series
    Posted in the Algebra Forum
    Replies: 4
    Last Post: October 21st 2010, 09:57 AM
  2. Convergence/Divergence of a series.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 10th 2010, 11:22 AM
  3. Help with series convergence or divergence
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 12th 2009, 05:04 PM
  4. Convergence/Divergence of a series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 4th 2008, 06:41 PM
  5. Series(convergence divergence)
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: November 8th 2005, 04:28 PM

/mathhelpforum @mathhelpforum