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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.

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To simplify matters a little, I can do with an estimate of the form $\prod (1 + e^{|x|^n - j^{1/(n-1)}}) \le Ce^{C|x|^m}$ for some $m$. – flavio Mar 18 '13 at 14:58
up vote 1 down vote accepted

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 $$ \exp(c_n+c_n|x|^{n^2}),\qquad c_n=2^n(1+(n-1)!). $$

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Thank you so much! Much appreciated. – flavio Mar 21 '13 at 10:38

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