For any integer $n\ge 3$, let $P(x)=\sum\limits\_{i(=2k)\ge 0}^{n}\binom{n}{2k}(1-x)^k$, $Q(x)=\sum\limits\_{i(=2k+1)\ge 0}^{n}\binom{n}{2k+1}(1-x)^{k+1}$. Define $\frac{P(x)}{Q(x)}\equiv\sum\limits\_{i=0}^{\infty}c\_ix^i $. I like to know $c\_0> c\_1> c\_2>\cdots$, but I don't know how to show it.



Decreasing coefficients?
The polynomials $P\_n$ and $Q\_n$ can be written as
$$
P\_n(x)=(1+\sqrt{1-x})^n+(1-\sqrt{1-x})^n,
\qquad
Q\_n(x)=\sqrt{1-x}\bigl((1+\sqrt{1-x})^n-(1-\sqrt{1-x})^n\bigr).
$$
In particular, they are both solutions to the linear difference equation
$P\_{n+1}(x)=2P\_n(x)-xP\_{n-1}(x)$ implying that their quotient
$f\_n(x)=P\_n(x)/Q\_n(x)$ satisfies the nonlinear recursion
$$
f\_{n+1}(x)=\frac{1+f\_n(x)}{1+(1-x)f\_n(x)}
$$
(already observed experimentally by Martin Rubey).


The property saying that the coefficients of the power series
$f(x)=c\_0+c\_1x+c\_2x^2+\dots$ satisfy $c\_0\ge c\_1\ge c\_2\ge\dots$
can be rephrased as $c\_0-(1-x)f(x)\ge0$, where the record
$a\_0+a\_1x+a\_2x^2+\dots\ge0$ means that all $a\_i\ge0$. This makes
the original problem equivalent to $1-(1-x)f\_n(x)\ge0$ for all $n$.
This follows by induction on $n$. The base of induction is clear
verification (already done by the author for a few first $n$).
Assuming that it is true for a given $n$ and using that the
non-negative expansion $1-(1-x)f\_n(x)$ has the vanishing constant term,
we conclude that the (formal) power series
$$
\frac1{1-\frac12\bigl(1-(1-x)f\_n(x)\bigr)}
=1+\sum\_{k=1}^\infty\frac1{2^k}\bigl(1-(1-x)f\_n(x)\bigr)^k
$$
is non-negative. It remains to apply the above recursion for $f\_n(x)$:
$$
1-(1-x)f\_{n+1}(x)
=\frac x{1+(1-x)f\_n(x)}
=\frac x2\cdot\frac1{1-\frac12\bigl(1-(1-x)f\_n(x)\bigr)},
$$
and the desired property follows.


