Find whether : $\sum_{ I \subset \mathbb{N}} e^{-\sqrt{S(I)}}$ converges

To make things precise, when $|I| = \infty$, we redefine the value of $S(I)$ as $+\infty$ and $e^{-\sqrt{S(I)}} = 0$.

After this change, the summand inside the sum $\sum\limits_{I\subset \mathbb{N}} e^{-\sqrt{S(I)}}$ will be non-zero for countably many $I$. Since all summands are non-negative, the sum is well defined and takes value in $[0,\infty]$. Furthermore, we can compute it by enumerating those $I$ with $|I| < \infty$ in arbitrary order and get the same result. As a result,

$$\begin{align}\sum_{I\subset \mathbb{N}} e^{-\sqrt{S(I)}} \stackrel{def}{=} \sum_{I\subset \mathbb{N}, |I| < \infty} e^{-\sqrt{S(I)}} &= 2\sum_{I\subset \mathbb{Z}_{+}, |I| < \infty} e^{-\sqrt{S(I)}} = 2\sum_{n=0}^\infty \sum_{I \subset \mathbb{Z}_{+}, S(I) = n} e^{-\sqrt{n}}\\ &= 2\sum_{n=0}^\infty q(n) e^{-\sqrt{n}} \end{align} $$

where $q(n) = | \{ I \subset \mathbb{Z}_{+} : S(I) = n \} |$ is the number of partitions of integer $n$ into distinct parts.

The OGF of $q(n)$ equals to

$$\sum_{n=0}^\infty q(n) z^n = \prod_{k=1}^\infty ( 1 + z^k )$$

The closed form of $q(n)$ is not known. However, we do know for large $n$,${}^{\color{blue}{[1]}}$

$$q(n) \sim \frac{3^{3/4}}{12 n^{3/4}} \exp\left(\pi\sqrt{\frac{n}{3}}\right) $$

Since $\alpha \stackrel{def}{=} \frac{\pi}{\sqrt{3}} - 1 > 0$, the sub-sum over those $I$ with $S(I) = n$ blows up like $e^{\alpha\sqrt{n}}$. From this, we can deduce

$$\sum_{I\subset \mathbb{N}} e^{-\sqrt{S(I)}} = \infty$$

Refs

• $\color{blue}{[1]}$ - Philippe Flajolet, Robert Sedgewick Analytic Combinatorics, Cambridge University Press; (1st ed., 2009). Formula found at VIII.6 Saddle-point asymptotics / Integer partitions.