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Calculate this sum

  1. Feb 8, 2012 #1
    Hello, I want to calculate the sum of this series:
    [tex]\sum_{n=1}^\infty \frac{3^n+1}{4^n+2}[/tex],
    I know the series converges, I only found this useful way to factor the denominator:
    [tex] 4^n+2=2^{2n}+2=4^n(1+2^{-2n+1})[/tex],
    now i have: [tex]\frac{3^n+1}{4^n+2}=(\frac{3^n}{4^n}+\frac{1}{4^n})(\frac{1}{1+2^{-2n+1}})[/tex],
    i can calculate [tex]\sum_{n=1}^\infty \frac{3^n}{4^n}+ \sum_{n=1}^\infty \frac{1}{4^n}[/tex]but what should be done with
  2. jcsd
  3. Feb 8, 2012 #2


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    Gold Member

    I'm not sure you can do the series you asked about, but you can get an upper bound if that helps by noticing that

    [tex] \frac{1}{1-\frac{3 x}{4}}+\frac{1}{1-\frac{x}{4}} [/tex]

    is a generating function for

    [tex] \frac{3^n+1}{4^n} [/tex]

    Maybe you could find a lower bound too and sandwich it.
  4. Feb 8, 2012 #3


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    By the way, this bound can be made really tight. For example, numerically we get

    [tex]\sum _{n=1}^{\infty } \frac{1+3^n}{2+4^n} = 2.9142434104[/tex]

    Now since the majority of the difference between the series I proposed as an upper bound and the one you care about happens in the lower terms, evaluate a few of the lower terms exactly and do the remaining terms using the approximation.

    [tex] \sum _{n=1}^{10 } \frac{1+3^n}{2+4^n} + \left(\frac{1}{1-\frac{3 }{4}}+\frac{1}{1-\frac{1}{4}} - \sum _{n=0}^{10 } \frac{1+3^n}{4^n} \right) = \frac{1747777419267069422319725824985}{599736246509717885804751618048}[/tex]
    [tex] = 2.914243435 [/tex]

    So the difference is on the order of [itex] 10^{-8} [/itex]
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