How fast does $\Sigma_{n=1}^{\infty} (1- \exp(-x))^{n^2} $ grow?

Question

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.

Answer 1

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 $x \to \infty$.

Answer 2

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.

Answer 3

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

$$ \sum_{n=0}^{\infty} (1 - e^{-x})^{n^2} \sim \frac{\sqrt{\pi}}{2} e^{x/2} \quad \text{as } x \to \infty. $$