Results 1 to 10 of 10

Math Help - series convergence

  1. #1
    dim
    dim is offline
    Newbie
    Joined
    Nov 2008
    Posts
    5

    series convergence

    Series a_n converges and a_n>=0. Does series e^{a_n}-1 converge as well?

    I suppose sum {e^{a_n}} < e^{sum{a_n}} => the sequence of partial sums converges and the series should converge. But lim e^{a_n} /= 0!What is wrong ?

    The same question about convergence for series n*a_{n^2}.

    Thanks for any help=)
    Last edited by dim; November 21st 2008 at 04:21 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by dim View Post
    Series a_n converges and a_n>=0. Does series e^{a_n}-1 converge as well?

    I suppose that sum(e^{a_n}) < e^(sum{a_n}) => the sequence of partial sums converges and the series should converge. But lim e^{a_n} /=0! What is wrong ?

    The same question about convergence for series n*a_{n^2}.

    Thanks for any help=)
    e^{a_n}-1=1+a_n+\frac{a_n^2}{2}+\cdots-1=a_n+\frac{a_n^2}{2}+\cdots=\sum_{N=1}^{\infty}\f  rac{a_n^N}{N!}

    Can you go from there?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    dim
    dim is offline
    Newbie
    Joined
    Nov 2008
    Posts
    5
    Indeed I cannot use series expansion yet! Is there any other way?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by dim View Post
    Indeed I cannot use series expansion yet! Is there any other way?
    Yeah I thought of another way. Consider \lim_{n\to\infty}\frac{e^{a_n}-1}{a_n}. Now let \varphi=a_n so because the series \sum{a_n} converges we know that a_n is a null sequence (it converges to zero). So we see that as n\to\infty\implies{\varphi\to{0}}. So our limit becomes \lim_{\varphi\to{0}}\frac{e^{\varphi}-1}{\varphi}=1. Thus the two series \sum\left\{{e^{a_n}-1}\right\} and \sum{a_n} share convergence.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by dim View Post
    Series a_n converges and a_n>=0.
    The same question about convergence for series n*a_{n^2}.

    Thanks for any help=)
    Is there any more info about the second one? If you have some extra restrictions on the set you should say. I have come up with a solution, but it only applies if you have a few more restrictions, report back if this is the exact question.

    Consider \sum{na_{n^2}}, we know that a_n=0, now ASSUMING that a_n\in\downarrow and a_n\in\mathcal{C} we have that this series and the integral \int_{0}^{\infty}na_{n^2}dn share convergence. Now let \psi=n^2 then we have that

    \frac{1}{2}\int_0^{\infty}a_{\psi}d\psi, now since we are given that \sum{a_n} converges so does \frac{1}{2}\int_0^{\infty}a_{\psi}d\psi\quad\there  fore\sum{na_{n^2}}\text{ converges}\quad\blacksquare
    Follow Math Help Forum on Facebook and Google+

  6. #6
    dim
    dim is offline
    Newbie
    Joined
    Nov 2008
    Posts
    5
    Quote Originally Posted by Mathstud28 View Post
    Is there any more info about the second one?
    No other restrictions, only that a_n is positive and converges (series).
    I thought about finding \lim_{n\to\infty}\frac{(n+1)a_{(n+1)^2}}{na_{n^2}}. We know that  \lim_{n\to\infty}\frac{a_{n^2+1}}{a_{n^2}}=0 as series a_n converges. Can we say that \lim_{n\to\infty}\frac{a_{(n+1)^2}}{a_{n^2}}=0? If so we could probably use this test.

    P.S. And I see we can by applying chain multiplication \lim_{n\to\infty}\frac{a_{(n+1)^2}}{a_{n^2}}=\lim_  {n\to\infty}\frac{a_{n^2+1+2n}}{a_{n^2+1+2n-1}}*\frac{a_{n^2+1+2n-1}}{a_{n^2+1+2n-2}}*...*\frac{a_{n^2+1}}{a_{n^2}}=0
    Last edited by dim; November 21st 2008 at 02:22 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    dim
    dim is offline
    Newbie
    Joined
    Nov 2008
    Posts
    5
    Quote Originally Posted by Mathstud28 View Post
    Yeah I thought of another way. Consider \lim_{n\to\infty}\frac{e^{a_n}-1}{a_n}. Now let \varphi=a_n so because the series \sum{a_n} converges we know that a_n is a null sequence (it converges to zero). So we see that as n\to\infty\implies{\varphi\to{0}}. So our limit becomes \lim_{\varphi\to{0}}\frac{e^{\varphi}-1}{\varphi}=1. Thus the two series \sum\left\{{e^{a_n}-1}\right\} and \sum{a_n} share convergence.
    Thanks a lot! It is great!

    There is one more question I could not deal with .. I need to show that \sum{\frac{1}{nlnn}} is convergent, but I can not use integral test for convergence! If you have any idea, I will highly appreciate it.
    Last edited by dim; November 21st 2008 at 02:32 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by dim View Post
    No other restrictions, only that a_n is positive and converges (series).
    I thought about finding \lim_{n\to\infty}\frac{(n+1)a_{(n+1)^2}}{na_{n^2}}. We know that  \lim_{n\to\infty}\frac{a_{n^2+1}}{a_{n^2}}=0 as series a_n converges. Can we say that \lim_{n\to\infty}\frac{a_{(n+1)^2}}{a_{n^2}}=0? If so we could probably use this test.

    P.S. And I see we can by applying chain multiplication \lim_{n\to\infty}\frac{a_{(n+1)^2}}{a_{n^2}}=\lim_  {n\to\infty}\frac{a_{n^2+1+2n}}{a_{n^2+1+2n-1}}*\frac{a_{n^2+1+2n-1}}{a_{n^2+1+2n-2}}*...*\frac{a_{n^2+1}}{a_{n^2}}=0
    I do not have time now to do check this out thourougly but consider that \lim_{n\to\infty}\frac{a_{(n+1)^2}}{a_{n^2}} is the Ratio test for this series.
    Quote Originally Posted by dim View Post
    Thanks a lot! It is great!

    There is one more question I could not deal with .. I need to show that \sum{\frac{1}{nlnn}} is convergent, but I can not use integral test for convergence! If you have any idea, I will highly appreciate it.
    Use Cauchy's Condensation test. A series satisfying the same restrictions (save continuity) of the integral test converges iff \sum{2^na_{2^n}} converges, so now consider \frac{2^n}{2^n\ln\left(2^n\right)}. Use your laws of logarithims and this becomes the Harmonic Series, thus both this and your question series diverge by CC's test.
    Last edited by Mathstud28; November 21st 2008 at 01:04 PM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,413
    Thanks
    1852
    A series does not converge unless the sequence of numbers converges to 0. In particular, \sum e^{a_n}- 1 cannot converge if e^{a_n} does not converge to 1 which is the same as saying that a_n converges to 0.

    Of course, if a_n does converge to 0, it does not follow that [tex]\sum e^{a_n}-1[tex] converges.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by HallsofIvy View Post
    A series does not converge unless the sequence of numbers converges to 0. In particular, \sum e^{a_n}- 1 cannot converge if e^{a_n} does not converge to 1 which is the same as saying that a_n converges to 0.

    Of course, if a_n does converge to 0, it does not follow that [tex]\sum e^{a_n}-1[tex] converges.
    What does that have to do with anything?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: October 3rd 2011, 02:12 AM
  2. Replies: 2
    Last Post: May 1st 2010, 10:22 PM
  3. Replies: 4
    Last Post: December 1st 2009, 04:23 PM
  4. Replies: 7
    Last Post: October 12th 2009, 11:10 AM
  5. series convergence and radius of convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 15th 2008, 09:07 AM

Search Tags


/mathhelpforum @mathhelpforum