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I want to find a specific range of $\alpha$ formula as follows.

$$\alpha =\frac{1+\frac{{{x}^{2}}}{3!}+\frac{{{x}^{4}}}{5!}++\cdots+{\frac {x^{2n}}{(2n+1)!}}}{1+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{4}}}{4!}+\cdots +{\frac {x^{2n}}{(2n)!}}}$$

We know that 0 < $\alpha$ < 1. But, I want to narrow this range or make it more specific value as much as possible? If it is possible, could you please suggest me some ways to do it without depending on the variable $n$ ($n\to \infty $)?

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  • $\begingroup$ @metamorphy, can you suggest an answer without depending on the 'n' variable (consider $n\to \infty $)? $\endgroup$
    – Karim
    Commented Apr 20, 2020 at 10:00
  • $\begingroup$ Numerically, it seems that $$\frac{{\tanh x}}{{x}} \le \alpha (x) \le \frac{{x^2 + 6}}{{3x^2 + 6}} \le 1.$$ $\endgroup$
    – Gary
    Commented Apr 20, 2020 at 10:10
  • $\begingroup$ Yes, I want to find a minimum value (a constant) of this formula. Is it possible? How could we derive it? can you suggest? $\endgroup$
    – Karim
    Commented Apr 20, 2020 at 10:11
  • $\begingroup$ It seems that for any fixed $x$ it is a monotonically decreasing sequence with respect to $n$. I do not have a proof though. $\endgroup$
    – Gary
    Commented Apr 20, 2020 at 10:19

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