Page 1 of 2 12 LastLast
Results 1 to 15 of 21

Thread: Test the convergence of the series

  1. #1
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    Test the convergence of the series

    Question :

    Test the convergence of the series

    $\displaystyle
    \frac{x^2}{3 \sqrt{2}} + \frac{x^4}{4 \sqrt{3}} + \frac{x^6}{5 \sqrt{4}} + ...
    $ for all x
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by zorro View Post
    Question :

    Test the convergence of the series

    $\displaystyle
    \frac{x^2}{3 \sqrt{2}} + \frac{x^4}{4 \sqrt{3}} + \frac{x^6}{5 \sqrt{4}} + ...
    $ for all x
    The general term is $\displaystyle \frac{x^{2n}}{(n + 2) \sqrt{n + 1}}$. Now choose an appropriate test from the list of tests you have been taught and apply it.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    I am stuck here

    Quote Originally Posted by mr fantastic View Post
    The general term is $\displaystyle \frac{x^{2n}}{(n + 2) \sqrt{n + 1}}$. Now choose an appropriate test from the list of tests you have been taught and apply it.


    I am using the ratio test to test the series ...

    $\displaystyle \lim_{n \to \infty} \frac{a_n + 1 }{a_n} = \lim_{n \to \infty} \frac{x^{2(n+1)}}{(n + 3) \sqrt{n + 2}} . \frac{(n + 2) \sqrt{n + 1}}{x^{2n}}$


    = $\displaystyle \lim_{n \to \infty} \frac{x^{2n} . x^2}{(n + 3) \sqrt{n + 2}} . \frac{(n + 2) \sqrt{n + 1}}{x^{2n}}$



    = $\displaystyle \lim_{n \to \infty} \frac{x^2}{(n + 3) \sqrt{n + 2}} . \frac{(n + 2) \sqrt{n + 1}}{1}$


    = $\displaystyle \lim_{n \to \infty} \frac{x^2 \left[(n + 2) \sqrt{n + 1} \right]}{(n + 3) \sqrt{n + 2}} $

    I am stuck here!!!!!!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    try to use root test.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    I am using the root test

    Quote Originally Posted by Krizalid View Post
    try to use root test.

    I am using the root test and have got this answer where i am stuck right now
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by zorro View Post
    I am using the ratio test to test the series ...

    $\displaystyle \lim_{n \to \infty} \frac{a_n + 1 }{a_n} = \lim_{n \to \infty} \frac{x^{2(n+1)}}{(n + 3) \sqrt{n + 2}} . \frac{(n + 2) \sqrt{n + 1}}{x^{2n}}$


    = $\displaystyle \lim_{n \to \infty} \frac{x^{2n} . x^2}{(n + 3) \sqrt{n + 2}} . \frac{(n + 2) \sqrt{n + 1}}{x^{2n}}$



    = $\displaystyle \lim_{n \to \infty} \frac{x^2}{(n + 3) \sqrt{n + 2}} . \frac{(n + 2) \sqrt{n + 1}}{1}$


    = $\displaystyle \lim_{n \to \infty} \frac{x^2 \left[(n + 2) \sqrt{n + 1} \right]}{(n + 3) \sqrt{n + 2}} $

    I am stuck here!!!!!!
    From your last step, you can pull out the $\displaystyle x^2$ (because it has no $\displaystyle n$ in it), so

    $\displaystyle \lim_{n \to \infty} \frac{x^2(n + 2) \sqrt{n + 1}}{(n + 3) \sqrt{n + 2}} = x^2\lim_{n\to\infty}\frac{\sqrt{(n+2)^2(n+1)}}{\sq rt{(n+3)^2(n+2)}}=x^2\lim_{n\to\infty}\frac{\sqrt{ n^3+5n^2+8n+4}}{\sqrt{n^3+8n^2+21n+18}}$

    Spoiler:
    Now pull the $\displaystyle n^3$ out of the square root to get:

    $\displaystyle x^2\lim_{n\to\infty}\frac{n^{3/2}\sqrt{1+5/n+8/n^2+4/n^3}}{n^{3/2}\sqrt{1+8/n+21/n^2+18/n^3}}$

    Now you can see that taking $\displaystyle n\to\infty$ gives you a limit of $\displaystyle 1$, (i.e. the whole thing equals $\displaystyle x^2$), so the series converges when $\displaystyle |x^2|<1$, so when $\displaystyle |x|<1$. You also need to check for convergence at the endpoints ($\displaystyle 1$ and $\displaystyle -1$).
    Spoiler:
    For $\displaystyle -1$, use the alternating series test. For $\displaystyle +1$, compare it to $\displaystyle \sum\frac{1}{n^{3/2}}$.
    Spoiler:
    Just kidding; two nested spoilers is enough.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    Alternating series for -1

    Quote Originally Posted by redsoxfan325 View Post
    From your last step, you can pull out the $\displaystyle x^2$ (because it has no $\displaystyle n$ in it), so

    $\displaystyle \lim_{n \to \infty} \frac{x^2(n + 2) \sqrt{n + 1}}{(n + 3) \sqrt{n + 2}} = x^2\lim_{n\to\infty}\frac{\sqrt{(n+2)^2(n+1)}}{\sq rt{(n+3)^2(n+2)}}=x^2\lim_{n\to\infty}\frac{\sqrt{ n^3+5n^2+8n+4}}{\sqrt{n^3+8n^2+21n+18}}$

    Spoiler:
    Now pull the $\displaystyle n^3$ out of the square root to get:

    $\displaystyle x^2\lim_{n\to\infty}\frac{n^{3/2}\sqrt{1+5/n+8/n^2+4/n^3}}{n^{3/2}\sqrt{1+8/n+21/n^2+18/n^3}}$

    Now you can see that taking $\displaystyle n\to\infty$ gives you a limit of $\displaystyle 1$, (i.e. the whole thing equals $\displaystyle x^2$), so the series converges when $\displaystyle |x^2|<1$, so when $\displaystyle |x|<1$. You also need to check for convergence at the endpoints ($\displaystyle 1$ and $\displaystyle -1$).
    Spoiler:
    For $\displaystyle -1$, use the alternating series test. For $\displaystyle +1$, compare it to $\displaystyle \sum\frac{1}{n^{3/2}}$.
    Spoiler:
    Just kidding; two nested spoilers is enough.



    Thanks for the reply....i am stuck again ....I know it is a silly question but could u please tell me on what should i use the alternate series test for

    I know i need to use for -1 but which equation should i use
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by zorro View Post
    Thanks for the reply....i am stuck again ....I know it is a silly question but could u please tell me on what should i use the alternate series test for

    I know i need to use for -1 but which equation should i use
    Since your series involves even powers of x, x = 1 and x = -1 give the same series. The alternating series test is not relevant because the terms do not alternate when x = -1. The comparison test, implied in the spoiler of an earlier reply, is all that is required.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    My question wasnt that

    Quote Originally Posted by mr fantastic View Post
    Since your series involves even powers of x, x = 1 and x = -1 give the same series. The alternating series test is not relevant because the terms do not alternate when x = -1. The comparison test, implied in the spoiler of an earlier reply, is all that is required.
    Mr fantastic i wanted to know where should i put in the comparison test is it to be put in the eq $\displaystyle \frac{x^2n}{n + 2 \sqrt{n + 1}}$ or in the result $\displaystyle x^2 $
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by zorro View Post
    Mr fantastic i wanted to know where should i put in the comparison test is it to be put in the eq $\displaystyle \frac{x^2n}{n + 2 \sqrt{n + 1}}$ or in the result $\displaystyle x^2 $
    Substitute x = 1 into the original series. Then use the comparison test on the resulting series, as shown in the spoiler of an earlier reply. This will also establish what happens when x = -1 (because substituting x = -1 gives the same series).
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    Is this correct?

    Quote Originally Posted by mr fantastic View Post
    Substitute x = 1 into the original series. Then use the comparison test on the resulting series, as shown in the spoiler of an earlier reply. This will also establish what happens when x = -1 (because substituting x = -1 gives the same series).

    Substitute x = 1 in the original series we get

    $\displaystyle
    \frac{(-1)^2}{3 \sqrt{2}} + \frac{(-1)^4}{4 \sqrt{3}} + \frac{(-1)^6}{5 \sqrt{4}}
    $

    $\displaystyle
    u_n = \frac{(-1)^2n}{(n + 2) \sqrt{n + 1}}
    $ and $\displaystyle
    v_n = \frac{1}{n^{\frac{3}{2}}} $


    $\displaystyle
    \lim_{n \to \infty} \frac{u_n}{v_n} = \lim_{n \to \infty} \frac{\frac{(-1)^2n}{(n + 2) \sqrt{n + 1}} }{\frac{1}{n^{\frac{3}{2}}}}$ = $\displaystyle \lim_{n \to \infty} \frac{(-1)^2n}{(n + 2) \sqrt{n + 1}} . n^{\frac{3}{2}}$

    Is this correct????????
    Last edited by zorro; Dec 7th 2009 at 11:42 AM.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Sorry, my bad. I didn't notice it was all even powers. In that case, -1 and 1 are the same.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    Is the answer correct ?

    Quote Originally Posted by redsoxfan325 View Post
    Sorry, my bad. I didn't notice it was all even powers. In that case, -1 and 1 are the same.
    In my last post i have done some step could u please check and let me know if i have done it correctly or no .....and what else do i need to do after this
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by zorro View Post
    Substitute x = 1 in the original series we get

    $\displaystyle
    \frac{(-1)^2}{3 \sqrt{2}} + \frac{(-1)^4}{4 \sqrt{3}} + \frac{(-1)^6}{5 \sqrt{4}}
    $

    $\displaystyle
    u_n = \frac{(-1)^{2n}}{(n + 2) \sqrt{n + 1}}
    $ and $\displaystyle
    v_n = \frac{1}{n^{\frac{3}{2}}} $


    $\displaystyle
    \lim_{n \to \infty} \frac{u_n}{v_n} = \lim_{n \to \infty} \frac{\frac{(-1)^{2n}}{(n + 2) \sqrt{n + 1}} }{\frac{1}{n^{\frac{3}{2}}}}$ = $\displaystyle \lim_{n \to \infty} \frac{(-1)^{2n}}{(n + 2) \sqrt{n + 1}} . n^{\frac{3}{2}}$

    Is this correct????????
    So far, yes. Now calculate the limit. You should find that it equals 1, which tells you...
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    I am stuck inthe limit

    Quote Originally Posted by redsoxfan325 View Post
    So far, yes. Now calculate the limit. You should find that it equals 1, which tells you...

    Could u please tell me how could i get rid of $\displaystyle (-1)^2n$
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Test the series for convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Aug 15th 2010, 04:29 PM
  2. Replies: 4
    Last Post: Feb 10th 2010, 08:08 AM
  3. Test the convergence of the series
    Posted in the Calculus Forum
    Replies: 9
    Last Post: Dec 23rd 2009, 02:06 AM
  4. test for convergence of series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 16th 2009, 09:21 AM
  5. Test the Convergence of the series
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Dec 23rd 2008, 03:39 PM

Search Tags


/mathhelpforum @mathhelpforum