Estimate of infinite product I'm wondering how one can obtain an estimate such as
$$
\prod _{j=1}^\infty (1 + e^{|x|^n - j^{\frac{1}{n-1}}}) \le Ce^{C|x|^{n(n-1)}}, 
$$
for some constant $C>0$ where $n$ is a positive integer and $x\in \mathbb{R}$.
I'm familiar with estimates such as 
$$
\prod (1 + a_j) \le e^{\sum a_j}
$$ 
for summable sequences $(a_j)$ but not sure how to arrive at the above inequality. 
 A: For every $j\geqslant1$, $a\gt0$ and $z\geqslant0$,
$$
\log(1+\mathrm e^{z-j^{1/a}})\leqslant\int_{j-1}^j\log(1+\mathrm e^{z-x^{1/a}})\mathrm dx,
$$
hence the infinite product you are interested in is $\leqslant\mathrm e^L$ with
$$
L=\int_{0}^\infty\log(1+\mathrm e^{z-x^{1/a}})\mathrm dx=\int_{0}^\infty\log(1+\mathrm e^{z-x})ax^{a-1}\mathrm dx.
$$
An integration by parts yields
$$
L=\int_{0}^\infty x^a\frac{\mathrm e^z}{\mathrm e^z+\mathrm e^x}\mathrm dx.
$$
If $x\leqslant2z$, the ratio in the integral is $\leqslant1$. If $x\geqslant2z$, it is $\leqslant1/(1+\mathrm e^{x/2})$. Thus,
$$
L\leqslant\int_0^{2z}x^a\mathrm dx+\int_{2z}^{+\infty}x^a\frac{\mathrm dx}{1+\mathrm e^{x/2}}\leqslant(2z)^{a+1}+(2z)^{a+1}\int_1^{+\infty}x^a\frac{\mathrm dx}{1+\mathrm e^{zx}}.
$$
For every $z\geqslant1$, a simplified version of the upper bound is
$$
L\leqslant(2z)^{a+1}+(2z)^{a+1}\Gamma(a+1),
$$
hence, for $a=n-1$ and $z=|x|^n$, the original product is at most
$$
\exp(c_n+c_n|x|^{n^2}),\qquad c_n=2^n(1+(n-1)!).
$$
