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.

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