Find whether : $\sum_{ I \subset \mathbb{N}} e^{-\sqrt{S(I)}}$ converges 
Does the following sum converges : $$\sum_{ I \subset \mathbb{N}} e^{-\sqrt{S(I)}}$$
where : $$ S(I) =\sum_{ i \in I} i$$

I don’t know how to approach this problem. Nevertheless, maybe it’s possible to have some intuition about the problem.
If $\mid I \mid = n$ then we know that the $S(I)$ which are going to give an important weight to the sum are in $O(n^2)$. Hence maybe luckily :
$$\sum_{I \subset \mathbb{N}, S(I) = O(\mid I \mid ^2)} e^{-\sqrt{S(I)}}$$
has the same nature of our sum.
In this case the square root makes it easy and the sum converges.
Yet it’s possible that this intuition is false since there are a lot of $S(I) \ne O(\mid I \mid ^2)$ (infinitely many actually) so it might give a lot of weight to the sum... I don’t really know.
Note that in order that $S(I)$ makes sens, we have : $I < \infty$.
 A: 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.

A: That sum, or just any sum with more than countably many positive summands cannot converge:
Let  $X$ be a set and $f\colon X\to\Bbb R$ a map such that $f(x)> 0$ fore all $x\in X$. Assume that  $\sum_{x\in X}f(x)$ converges, say $\sum_{x\in X}f(x)=S\in \Bbb R$. Then for $Y\subseteq X$, clearly $\sum_{x\in Y}f(x)\le \sum _{x\in X}f(x)$. 
For $n\in \Bbb N$, let $X_n=\{\,x\in X\mid f(x)>\frac 1n\,\}$. Then $$M\ge \sum_{x\in X_n}f(x)\ge \sum_{x\in X_n}\frac 1n=\frac 1n|X_n|$$
and in particular, $X_n$ must be finite. As $X=\bigcup_{n\in\Bbb N}X_n$, we conclude that $X$ is countable.
