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Evaluate the sum $$\frac{1}{3} + \frac{1}{3^{1+\frac{1}{2}}}+\frac{1}{3^{1+\frac{1}{2}+\frac{1}{3}}}+\cdots$$

It seems that $1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n}$ approaches $\ln n$ as $n\to \infty$, but I'm not sure if this is useful. Also, $3^{\ln n} =e^{\ln n\cdot \ln 3}= n^{\ln 3}$, but I'm also not sure how this is useful.

edit: I know how to prove that it converges, but I was wondering if there was a closed form for this sum.

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  • $\begingroup$ WolframAlpha doesn't give an exact answer. Which is not in any way conclusive, but it probably means that if an answer exists, it's difficult to either find or describe. $\endgroup$ – Arthur Dec 30 '19 at 0:25
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    $\begingroup$ From where did you get this series? Wasn't the question wasn't regarding the convergence? $\endgroup$ – Zacky Dec 30 '19 at 0:38
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    $\begingroup$ We expect something like $1/(\ln 3 - 1)$, which is about $10$. It would be a miracle if a closed formula for the exact answer exists. $\endgroup$ – WhatsUp Dec 30 '19 at 0:51
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    $\begingroup$ @frank That's really useful context that would be good to include in the main post. Helping to answer an exercise that is expected to have a solution, and helping someone satisfy their personal curiosity are two completely different things. Both are welcome on this site, but they require different approaches. $\endgroup$ – Arthur Dec 30 '19 at 0:54
  • $\begingroup$ Your statement about $H_n=1+1/2+1/3+\cdots+1/n$ approaching $\log n$ is true in one sense: $H_n/\log n\to 1.$ But $H_n-\log n\to\gamma$ the Euler-Mascheroni constant, which is not zero. en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant $\endgroup$ – Thomas Andrews Dec 30 '19 at 2:26
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We can certainly impose some bounds on the value of the sum, via the asymptotic expansion $$H_n \sim \log n + \gamma + \frac{1}{2n} - \frac{1}{12n^2} + \cdots. \tag{1}$$ The crudest bound is to note for $0 < z < 1$ the sum $$f(z) = \sum_{n=1}^\infty z^{H_n}$$ is dominated by $$\begin{align*} f(z) &< z^\gamma \sum_{n=1}^\infty z^{\log n} \\ &= z^\gamma \sum_{n=1}^\infty e^{\log z \log n}\\ &= z^\gamma \sum_{n=1}^\infty n^{\log z} \\ &= z^\gamma \zeta(-\log z). \tag{2} \end{align*}$$ For $z = 1/3$, this gives us the comparison $$f(1/3) \approx 5.34863 < 5.688508.$$ More terms of the asymptotic expansion $(1)$ can be used to speed the computation. However, we must be careful since $(1)$ is centered around $n = \infty$, so convergence is poor for small $n$; we can compensate by computing the initial terms precisely, then using the asymptotic expanison for large $n$, resulting in rapid convergence.

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If this was something you just came up with, it is highly unlikely there is any obtainable closed form expression. Checking the number Wolfram|Alpha generates from sum (1/(3^(sum (1/k) from k=1 to n))) from n=1 to infinity in an inverse symbolic calculator, I did not find anything.

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Just out of curiosity, $$\sum _{n=1}^{\infty } 3^{-H_n}\approx 5.34863233867$$ which is close to $$10\frac{ {3^{1/3}}}{7-7^{3/4}}\approx 5.34863230401$$

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  • $\begingroup$ how did you come up with $10\cdot \dfrac{3^{1/3}}{7-7^{3/4}}$? $\endgroup$ – user733113 Dec 30 '19 at 15:40
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    $\begingroup$ @frank. In my former research group, a guy, just for the fun of it, made a small code to find this kind of approximation. Now, he is full professor and, for the fun ot it, I passed on the phone the decimal representation of the infinite sum with a lot of digits. In fact, I suspect that he looked for $3$ and $10-3=7$. $\endgroup$ – Claude Leibovici Dec 30 '19 at 15:46
  • $\begingroup$ That's incredible :P $\endgroup$ – Mr Pie Dec 31 '19 at 8:42
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A very direct and simple way of getting quite close to the sum value is replacing the sum with an integral and using the simplest approximation for n-th Harmonic number

$$H(n) \approx \ln(n) + \gamma $$

$$ \int\limits_{x=1}^{+\infty} \frac{1}{3^{\ln(x)+\gamma}} dx = \frac{1}{3^{\gamma}(\ln(3)-1)}$$

This value is about $5.3785$ while the sum is really around $5.3486$.

(It is some task to prove why this is a good approximation, but not impossible one.)

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