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Math Help - convergence question

  1. #1
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    convergence question

    \sum _{h=0}^\infty \frac{h}{2^h}
    by what method it converges?
    to what it converges?
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: convergence question

    Quote Originally Posted by transgalactic View Post
    \sum _{h=0}^\infty \frac{h}{2^h}
    by what method it converges?
    to what it converges?
    Hi transgalactic!

    The standard method I know is to define f(x) = \sum_{h=0}^\infty {h x^h}.
    Then your series corresponds to f(1/2).

    If you integrate f(x) you should be able to find its limit (using the formula for a geometric series).
    If you differentiate the result, you find f(x).
    Substituting x=1/2 gives you your answer.


    Alternatively, you can simply write out a number of terms.
    Rearrange the terms in a triangle.
    Find the sums of each row in the triangle.
    And sum the results.
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  3. #3
    Member anonimnystefy's Avatar
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    Re: convergence question

    hi transgalactic

    here are some tests for determining whether series are convergent or not: Convergence Tests -- from Wolfram MathWorld

    EDIT: if it were me i would either use the ratio test (Ratio Test -- from Wolfram MathWorld)
    Last edited by anonimnystefy; December 31st 2011 at 08:34 AM.
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  4. #4
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    Re: convergence question

    Root test is faster.
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  5. #5
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    Re: convergence question

    i did the a_n+1/a_n test
    and got 1/2

    so it converges to 1/2
    ?
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  6. #6
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    Re: convergence question

    Converges but that's not the value.
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  7. #7
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    Re: convergence question

    to what it converges?
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  8. #8
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    Re: convergence question

    Quote Originally Posted by transgalactic View Post
    to what it converges?
    \begin{gathered}  \sum\limits_{k = 0}^\infty  {x^k }  = \frac{1}{{\left( {1 - x} \right)}},\;|x| < 1 \hfill \\  \sum\limits_{k = 0}^\infty  {kx^{k - 1} }  = \frac{1}{{\left( {1 - x} \right)^2 }} \hfill \\  \sum\limits_{k = 0}^\infty  {kx^k }  = \frac{x}{{\left( {1 - x} \right)^2 }} \hfill \\ \end{gathered}
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  9. #9
    MHF Contributor FernandoRevilla's Avatar
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    Re: convergence question

    Quote Originally Posted by transgalactic View Post
    to what it converges?
    Another way: in general, \sum_{n=1}^{+\infty}nr^n converges if |r|<1 (use for example the ratio test). Denote S the sum of the series, we have:

    (i)\;\;S=r+2r^2+3r^3+\ldots \quad (ii)\;\; rS=r^2+2r^3+3r^4+\ldots

    This implies S-rS=r+r^2+r^3+\ldots=\frac{1}{1-r}-1=\frac{r}{1-r} as a consequence S=\frac{r}{(1-r)^2} . For your case, substitute r=1/2 .
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