TITLE: Let $\sum_{n=1}^\infty a_n<\infty$. Then $\prod_{n=1}^\infty(1+a_n x)$ is subexponential in $x$.
QUESTION [5 upvotes]: In the process of trying to prove an estimate precluding blowup of a PDE, I have come across this problem. Suppose we are given an absolutely convergent sum, $\sum_{n=1}^\infty a_n<\infty$, with $a_n\ge0$. I claim that the function
$$
F(x) = \prod_{n=1}^\infty (1+a_n x)
$$
is sub-exponential in $x$, that is,
$$
\lim_{x\to+\infty} F(x) e^{-cx}=0
$$
for all $c>0$. For $a_n$ non-terminating, $F(x)$ is of course super-polynomial. Furthermore, the bound $1+a_n x\le e^{a_n x}$ gives us
$$
F(x) = \prod_{n=1}^\infty (1+a_n x)\le e^{\sum_{n=1}^\infty a_n x},\qquad x\ge0
$$
so $F(x)$ grows at most exponentially. However, I have yet to find a better bound on $F(x)$. Can we prove that $F(x)$ is in fact sub-exponential?
As an example of such an $F$, we consider the identity
$$
\prod_{n=1}^\infty \left( 1+\frac{x}{n^2} \right) = \frac{\sinh(\pi\sqrt x)}{\pi\sqrt x}
$$ which can be derived from Euler's infinite product representation of the sine function.

REPLY [4 votes]: Given $c > 0$ choose $N$ so large that $\sum_{n=N+1}^\infty a_n < c$. Now use your estimate $1+a_n x\le e^{a_n x}$ only for the factors with $n > N$:
$$
 f(x)e^{-cx} \le \prod_{n=1}^N (1+a_n x) \cdot \exp\left( \sum_{n=N+1}^\infty a_n  x \right) e^{-cx}
= \frac{\prod_{n=1}^N (1+a_n x) }{\exp\left( (c -\sum_{n=N+1}^\infty a_n ) x \right)}
$$
and that converges to zero because the exponential function grows faster than any polynomial.