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Thread: Testing the convergence

  1. #1
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    Testing the convergence

    Test the convergence of the series
    $\displaystyle
    \frac{1}{2\sqrt{1}} + \frac{x^2}{3\sqrt{2}} + \frac{x^4}{4\sqrt{3}}+\frac{x^6}{5\sqrt{4}}+.... \infty

    $

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

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

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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...

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

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$

    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...

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

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

    Convergence tests - Wikipedia, the free encyclopedia

    The series to be compared are...

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

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

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



    Merry Christmas from Italy

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

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

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

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

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

    Kind regards

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