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Thread: convergence problem

  1. #1
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    convergence problem

    I'm trying to understand how to get to this equality step by step. I appreciate your help.


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  2. #2
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    Re: convergence problem

    What have you tried? Are you familiar with geometric progressions? What happens when you take the derivative term-by-term?
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  3. #3
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    Re: convergence problem

    Quote Originally Posted by SlipEternal View Post
    What have you tried? Are you familiar with geometric progressions? What happens when you take the derivative term-by-term?
    I meant geometric sums of the form:

    $$\sum_{k=0}^\infty r^k = \dfrac{1}{1-r}, |r|<1$$
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  4. #4
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    Re: convergence problem

    Yes, I know how to do that geometric progression and I can do it with 2^(-k). But I have also tried to differentiate without success. Could you please guide me? Thank you!
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  5. #5
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    Re: convergence problem

    Quote Originally Posted by mathfn View Post
    Yes, I know how to do that geometric progression and I can do it with 2^(-k). But I have also tried to differentiate without success. Could you please guide me? Thank you!
    As I recall the series is $\displaystyle \sum\limits_{k = 1}^\infty {\frac{k}{{{2^k}}}} $.
    Lets consider $\displaystyle y = \sum\limits_{k = 1}^\infty {x^k}=\frac{x}{1-x} $ where $|x|<1$
    Now
    $\displaystyle \frac{dy}{dx} = \sum\limits_{k = 1}^\infty {kx^{k-1}}=\frac{1}{(1-x)^2} $, thus $\displaystyle x\cdot \frac{dy}{dx} = \sum\limits_{k = 1}^\infty {kx^{k}}=\frac{x}{(1-x)^2} $
    So use $x=\dfrac{1}{2}$ to get $\displaystyle \sum\limits_{k = 1}^\infty {\frac{k}{{{2^k}}}}=~? $

    P.S. SEE HERE
    Last edited by Plato; Oct 14th 2018 at 06:54 AM. Reason: P.S.
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