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Math Help - Evaluate this sum

  1. #1
    Super Member redsoxfan325's Avatar
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    Evaluate this sum

    Is there any way to evaluate this sum exactly?

    \frac{2}{3^1}+\frac{2}{3^2}+\frac{2}{3^6}+\frac{2}  {3^{24}}+... = \sum_{n=1}^{\infty}\frac{2}{3^{n!}}

    If not, can someone give me a good approximation? Also, is there any way to know whether this is rational or irrational? How about algebraic or transcendental?
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  2. #2
    MHF Contributor

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    Quote Originally Posted by redsoxfan325 View Post
    Is there any way to evaluate this sum exactly?

    \frac{2}{3^1}+\frac{2}{3^2}+\frac{2}{3^6}+\frac{2}  {3^{24}}+... = \sum_{n=1}^{\infty}\frac{2}{3^{n!}}

    If not, can someone give me a good approximation? Also, is there any way to know whether this is rational or irrational? How about algebraic or transcendental?
    let S=\sum_{k=1}^{\infty} \frac{1}{3^{k!}}. for a fixed n let S_n=\sum_{k=1}^n \frac{1}{3^{k!}}=\frac{p}{q}. then: \left |S-\frac{p}{q} \right|=\sum_{k=n+1}^{\infty} \frac{1}{3^{k!}}< \sum_{k=(n+1)!}^{\infty} \frac{1}{3^k}=\frac{1}{3^{(n+1)! - 1}} \leq \frac{1}{q^n}, where q=3^{n!}. thus S is a Liouville number and so it's transcendental.
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  3. #3
    Junior Member Infophile's Avatar
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    Hello,

    I don't believe it's possible to find the exact value, but a good approximation : \color{red}\frac{65}{729}

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