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Math Help - A problem involving sums.

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    A problem involving sums.

    The following question refers to http://www.mathhelpforum.com/math-he...-162658-3.html- a post of Bruno J. to chiph588@ problem.

    I will be glad if someone could explain me why \sum_{n=0}^\infty \frac1{\alpha^{n-m}+1}=\sum_{n=1}^m \frac1{\alpha^{-n}+1}+F(1)=\sum_{n=1}^m \frac1{\alpha^{-n}+1}+\sum_{n=0}^\infty \frac{1}{\alpha^n+1}



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  2. #2
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    Just notice \sum _{n=0}^{\infty} \frac{1}{\alpha ^{n-m}+1}=\sum _{n=0}^{m} \frac{1}{\alpha ^{n-m}+1}+\sum _{n=m+1}^{\infty} \frac{1}{\alpha ^{n-m}+1} and now just make k=n-m which gives you the equality between the first and last, and the rest follows from the definition of F
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Jose27 View Post
    Just notice \sum _{n=0}^{\infty} \frac{1}{\alpha ^{n-m}+1}=\sum _{n=0}^{m} \frac{1}{\alpha ^{n-m}+1}+\sum _{n=m+1}^{\infty} \frac{1}{\alpha ^{n-m}+1} and now just make k=n-m which gives you the equality between the first and last, and the rest follows from the definition of F
    Alright... but... why the second sum starts from 0(to infinity)? (It should be 1, no?)
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  4. #4
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Alright... but... why the second sum starts from 0(to infinity)? (It should be 1, no?)
    I suggest you read very carefully what I wrote and what you're asking. Isn't the first term off too?
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Jose27 View Post
    I suggest you read very carefully what I wrote and what you're asking. Isn't the first term off too?
    Just don't get mad...

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