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Math Help - Problem 30

  1. #1
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    Problem 30

    Show that \sum_{n=1}^{\infty}nx^{n-1} = \frac{1}{(1-x)^2} for |x|<1 without differenciating the geometric power series.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let S_n=\sum_{k=1}^nkx^{k-1}
    We have S_n=1+2x+3x^2+\ldots +nx^{n-1} (1).
    Multiplying (1) with x we have
    xS_n=x+2x^2+3x^3+\ldots +(n-1)x^{n-1}+nx^n (2)
    Substracting (2) from (1) yields
    S_n(1-x)=1+x+x^2+\ldots +x^{n-1}-nx^n=\frac{1-x^n}{1-x}-nx^n\Rightarrow S_n=\frac{1-x^n}{(1-x)^2}-\frac{nx^n}{1-x}
    Then \sum_{n=1}^{\infty}nx^{n-1}=\lim_{n\to\infty}S_n=\frac{1}{(1-x)^2}
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    Since you answered so quickly, here is a second challenge. I am not sure if is a fair question to ask but I post it anyway.
    ---

    Having shown that \sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2}

    Can you prove the following?
    Let f(x) = \sum_{n=0}^{\infty} a_nx^n with radius of convergence R>0

    Then, f is differenciable on (-R,R) and furthermore f'(x) = \sum_{n=1}^{\infty} na_nx^{n-1}.

    Note: This works even if x is a complex number, but the proofs are completely anagolous.
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  4. #4
    Super Member Rebesques's Avatar
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    I am not sure if is a fair question to ask but I post it anyway.
    That's ok, we just apply the relevant theorem
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  5. #5
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    Quote Originally Posted by Rebesques View Post
    That's ok, we just apply the relevant theorem
    I am not sure what you mean. In my Analysis class we proved differenciation term-by-term by working backwards with the Fundamental Theorem of Calculus after proving integration term-by-term. And that was just Real fucntions.


    By my Complex Analysis book had a really nice approach to this proof. Without using more advandeced techiqnues. It was able to prove term-by-term differenciation by using the series posted above. It was a long but straightforward derivation. That is what I am asking to show.
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