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Math Help - Testing the convergence

  1. #1
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    Testing the convergence

    Test the convergence of the series
    <br />
\frac{1}{2\sqrt{1}} + \frac{x^2}{3\sqrt{2}} + \frac{x^4}{4\sqrt{3}}+\frac{x^6}{5\sqrt{4}}+....         \infty<br /> <br />

    It is given that x^2 < 1
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  2. #2
    MHF Contributor chisigma's Avatar
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    The series is...

    \sum_{n=0}^{\infty} a_{n}\cdot x^{2n}= \sum_{n=0}^{\infty}\frac{x^{2n}}{(n+2)\cdot \sqrt{n+1}} (1)

    Because is a_{n}<1 , \forall n and the series...

    \sum_{n=0}^{\infty} x^{2n} (2)

    ... converges for |x|<1, the series (1) and (2) have the same radious of convergence...

    Kind regards

    \chi \sigma
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  3. #3
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    Could u please explain me

    Quote Originally Posted by chisigma View Post
    The series is...

    \sum_{n=0}^{\infty} a_{n}\cdot x^{2n}= \sum_{n=0}^{\infty}\frac{x^{2n}}{(n+2)\cdot \sqrt{n+1}} (1)

    Because is a_{n}<1 , \forall n and the series...

    \sum_{n=0}^{\infty} x^{2n} (2)

    ... converges for |x|<1, the series (1) and (2) have the same radious of convergence...

    Kind regards

    \chi \sigma

    thanks for ur reply but i couldnt understand it could u please explain me ....
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  4. #4
    MHF Contributor chisigma's Avatar
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    In this case we can prove the convergence of the series...

    \sum_{n=0}^{\infty} \frac{x^{2n}}{(n+2)\cdot \sqrt{n+1}} (1)

    ... using the 'comparison test'...

    Convergence tests - Wikipedia, the free encyclopedia

    The series to be compared are...

    \sum_{n=0}^{\infty} a_{n} , a_{n}= \frac{x^{2n}}{(n+2)\cdot \sqrt{n+1}}

    \sum_{n=0}^{\infty} b_{n} , b_{n}= x^{2n} (2)

    The series of the b_{n}= x^{2n} is 'geometric' and it converges for |x|<1. Since \forall n is |a_{n}|<|b_{n}| the convergence test extablishes that the (1) converges also for |x|<1...



    Merry Christmas from Italy

    \chi \sigma
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  5. #5
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    Quote Originally Posted by chisigma View Post
    The series is...

    \sum_{n=0}^{\infty} a_{n}\cdot x^{2n}= \sum_{n=0}^{\infty}\frac{x^{2n}}{(n+2)\cdot \sqrt{n+1}} (1)

    Because is a_{n}<1 , \forall n and the series...

    \sum_{n=0}^{\infty} x^{2n} (2)

    ... converges for |x|<1, the series (1) and (2) have the same radious of convergence...

    Kind regards

    \chi \sigma
    what is a_n ............i didnt get that part.........
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