<s>[INST] Find minimal value for product of the sums Find minimal value for product of the sums
$$\left( x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \right)
\left( \frac{1}{x}+\frac{1}{2!x^2}+\frac{1}{3!x^3}+\dots \right)
$$
Here what we did about solution. We know these series equals Taylor expansion of $(e^x-1)(e^{\frac{1}{x}}-1)$.  We tried to apply derivative test to find minimal value but it is hard to find extreme values from the equation we get.
Do you have any trick to find the minimal? Thank you for your help.
 [/INST] Hint. Set, for $x>0$,
$$
f(x)=(e^x-1)(e^{\frac{1}{x}}-1)=e^{x+\frac{1}{x}}-e^x-e^{\frac{1}{x}}+1
$$ giving
$$
f'(x)=\left(1-\frac1{x^2} \right)e^{x+\frac{1}{x}}-e^x+\frac1{x^2}e^{\frac{1}{x}}=\frac{e^{\frac{1}{x}}-e^{\frac{1}{x}+x}-e^x x^2+e^{\frac{1}{x}+x} x^2}{x^2}
$$ then observe that $f'(1)=0$, $f'(x)<0$ for $0<x<1$ and  $f'(x)>0$ for $x>1$.
The minimum is thus 

$$f(1)=(e-1)^2.$$

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