How fast does $\Sigma_{n=1}^{\infty} (1- \exp(-x))^{n^2} $ grow? Let $x$ be a positive real.
Define $\theta(x) = \Sigma_{n=1}^{\infty} (1- \exp(-x))^{n^2} $.
How fast does $\theta(x)$ grow ?
In other words what is a good asymptotic for it when $x$ is large ?
Maybe An integral representation for $\theta$ helps here.
Also similar functions occur in both algebra ( solving the quintic ) and number theory.
I assume $\theta$ has no nice inverse Laplace transform.
I also considered Trying contour integration , but also without succes.
 A: If $t = -\log(1 - e^{-x})$ then $t \to 0^+$ as $x \to \infty$ and
$$
\sum_{n=1}^{\infty} (1 - e^{-x})^{n^2} = \sum_{n=1}^{\infty} e^{n^2 \log(1 - e^{-x})} = \sum_{n=1}^{\infty} e^{-n^2 t}.
$$
Now
$$
-1 + \int_0^\infty e^{-x^2 t}\,dx \leq \sum_{n=1}^{\infty} e^{-n^2 t} \leq \int_0^\infty e^{-x^2 t}\,dx,
$$
and
$$
\int_0^\infty e^{-x^2 t}\,dx = \frac{1}{2}\sqrt{\frac{\pi}{t}},
$$
so
$$
\sum_{n=1}^{\infty} (1 - e^{-x})^{n^2} \sim \frac{1}{2}\sqrt{\frac{\pi}{t}} = \frac{1}{2}\sqrt{\frac{\pi}{-\log(1-e^{-x})}} \sim \frac{\sqrt{\pi}}{2} e^{x/2}
$$
as $x \to \infty$.
A: Let $(a_n)$ and $(s_n)$ be defined by
$$ a_n = \begin{cases} 1, & n \text{ is a square.} \\ 0, & \text{otherwise}. \end{cases}, \qquad s_n = a_0 + \cdots + a_n. $$
Then utilizing summation by parts,
$$ \sum_{n=0}^{\infty} x^{n^2}
= \sum_{n=0}^{\infty} a_n x^n
= \sum_{n=0}^{\infty} (s_n - s_{n-1}) x^n
= (1 - x) \sum_{n=0}^{\infty} s_n x^n.$$
By noticing that
$$ s_n = 1+\lfloor \sqrt{n} \rfloor \sim \sqrt{n} \sim (-1)^n \frac{\sqrt{\pi}}{2} \binom{-3/2}{n}, $$
it follows that as $x \to 1^{-}$
$$ (1 - x) \sum_{n=0}^{\infty} s_n x^n
\sim (1 - x) \sum_{n=0}^{\infty} (-1)^n \frac{\sqrt{\pi}}{2} \binom{-3/2}{n}
= \frac{\sqrt{\pi}}{2\sqrt{1-x}}. $$
(This can be thought as the power-series analogue of Cesaro-Stolz theorem.) Therefore
$$ \sum_{n=0}^{\infty} (1 - e^{-x})^{n^2} \sim \frac{\sqrt{\pi}}{2} e^{x/2} \quad \text{as } x \to \infty. $$
A: For a sane upper bound, we can use a geometric series. Consider $x\geq 0$. Then we have $$(1-\exp(-x))^{n^2}\leq (1-\exp(-x))^{n}$$ 
So
$$\sum_{n=1}^\infty(1-\exp(-x))^{n^2}\leq \sum_{n=1}^\infty(1-\exp(-x))^{n}$$
But the right hand side is geometric, so it's we know it's sum. That is,
$$\sum_{n=0}^\infty(1-\exp(-x))^{n^2} \leq \frac{1-e^{-x}}{e^{-x}}= e^x +1$$
Now this bound probably isn't very tight. The complex integration route is a clever idea, and might fruitful. I'll keep thinking about that. 
After some brief research, it looks like tighter bounds will depend on the Jacobi theta function.
