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Math Help - Test the convergence of the series

  1. #1
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    Test the convergence of the series

    Question :

    Test the convergence of the series

    <br />
\frac{x^2}{3 \sqrt{2}} + \frac{x^4}{4 \sqrt{3}} + \frac{x^6}{5 \sqrt{4}} + ...<br />
for all x
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  2. #2
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    Quote Originally Posted by zorro View Post
    Question :

    Test the convergence of the series

    <br />
\frac{x^2}{3 \sqrt{2}} + \frac{x^4}{4 \sqrt{3}} + \frac{x^6}{5 \sqrt{4}} + ...<br />
for all x
    The general term is \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.
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  3. #3
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    I am stuck here

    Quote Originally Posted by mr fantastic View Post
    The general term is \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 ...

    \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}}


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



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


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

    I am stuck here!!!!!!
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  4. #4
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    try to use root test.
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  5. #5
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    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
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  6. #6
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    Quote Originally Posted by zorro View Post
    I am using the ratio test to test the series ...

    \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}}


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



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


    = \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 x^2 (because it has no n in it), so

    \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 n^3 out of the square root to get:

    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 n\to\infty gives you a limit of 1, (i.e. the whole thing equals x^2), so the series converges when |x^2|<1, so when |x|<1. You also need to check for convergence at the endpoints ( 1 and -1).
    Spoiler:
    For -1, use the alternating series test. For +1, compare it to \sum\frac{1}{n^{3/2}}.
    Spoiler:
    Just kidding; two nested spoilers is enough.
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  7. #7
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    Alternating series for -1

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

    \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 n^3 out of the square root to get:

    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 n\to\infty gives you a limit of 1, (i.e. the whole thing equals x^2), so the series converges when |x^2|<1, so when |x|<1. You also need to check for convergence at the endpoints ( 1 and -1).
    Spoiler:
    For -1, use the alternating series test. For +1, compare it to \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
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  8. #8
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    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.
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  9. #9
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    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  \frac{x^2n}{n + 2 \sqrt{n + 1}} or in the result  x^2
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  10. #10
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    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  \frac{x^2n}{n + 2 \sqrt{n + 1}} or in the result  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).
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  11. #11
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    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

     <br />
\frac{(-1)^2}{3 \sqrt{2}} + \frac{(-1)^4}{4 \sqrt{3}} + \frac{(-1)^6}{5 \sqrt{4}}<br />

    <br />
u_n = \frac{(-1)^2n}{(n + 2) \sqrt{n + 1}} <br />
and <br />
v_n = \frac{1}{n^{\frac{3}{2}}}


    <br />
\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}}}} =  \lim_{n \to \infty} \frac{(-1)^2n}{(n + 2) \sqrt{n + 1}} . n^{\frac{3}{2}}

    Is this correct????????
    Last edited by zorro; December 7th 2009 at 11:42 AM.
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  12. #12
    Super Member redsoxfan325's Avatar
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    Sorry, my bad. I didn't notice it was all even powers. In that case, -1 and 1 are the same.
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  13. #13
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    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
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  14. #14
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by zorro View Post
    Substitute x = 1 in the original series we get

     <br />
\frac{(-1)^2}{3 \sqrt{2}} + \frac{(-1)^4}{4 \sqrt{3}} + \frac{(-1)^6}{5 \sqrt{4}}<br />

    <br />
u_n = \frac{(-1)^{2n}}{(n + 2) \sqrt{n + 1}} <br />
and <br />
v_n = \frac{1}{n^{\frac{3}{2}}}


    <br />
\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}}}} =  \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...
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  15. #15
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    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 (-1)^2n
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