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I Problems with wolfram alpha

  1. Sep 30, 2016 #1
    Hi, I'm reading a book of math and in one page says:
    $$\sum_{n=2}^{\infty }\frac{3^n-1}{4^n}\zeta (n+1)=\pi$$
    I tried to solve this,as I could not solved I requested to wolfram alpha and told me that this sum is
    approximately equal to 2.319125.
    https://www.wolframalpha.com/input/?i=sum+2+to+infinity+(3^k-1)/4^k+zeta(k+1)
    so i do not know if wolfram alpha is wrong or the book have a mistake and I like to know which is wrong.


    i tried some other values and the most proximate to pi is minus gamma, but I not quite shure and wolfram alpha calculation time expired u.u.

    so i do not know if this is true
    $$\sum_{n=2}^{\infty }\frac{3^n-1}{4^n}\zeta (n-\gamma )\approx \pi $$
     
  2. jcsd
  3. Sep 30, 2016 #2

    mfb

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  4. Sep 30, 2016 #3
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