# Does the sum $\frac{1}{3} + \frac{1}{3^{1+1/2}}+\cdots$ have a closed form?

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.

• 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. Dec 30, 2019 at 0:25
• From where did you get this series? Wasn't the question wasn't regarding the convergence? Dec 30, 2019 at 0:38
• 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. Dec 30, 2019 at 0:51
• @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. Dec 30, 2019 at 0:54
• 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 Dec 30, 2019 at 2:26

## 4 Answers

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.

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.

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

• how did you come up with $10\cdot \dfrac{3^{1/3}}{7-7^{3/4}}$?
– user733113
Dec 30, 2019 at 15:40
• @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 $10-3=7$. Dec 30, 2019 at 15:46
• That's incredible :P Dec 31, 2019 at 8:42

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.)