Show that certain sequence used in the proof of Wallis product formula is decreasing Define $I_0 := \pi/2, I_1 := 1$ and
$$ 
 I_{n+2} := \frac{n+1}{n+2} I_n.
$$
This sequence is monotone decreasing, which could be seen by recognizing that
$$
 I_n = \int_0^{\pi/2} \cos^n(x) \mathrm d x
$$
and $\cos^n(x) \ge \cos^{n+1}(x)$ for $x \in [0,\pi/2]$ (proof by induction and partial integration).
This sequence is used in the proof of Wallis product formula
$$
 \lim_{n \to \infty} \frac{2 \cdot 2}{1 \cdot 3} \cdot  \frac{4 \cdot 4}{3 \cdot 5} \cdots  \frac{2n\cdot 2n}{(2n-1)(2n+1)} = \frac{\pi}{2}
$$
by 
$$
 \frac{2 \cdot 2}{1 \cdot 3} \cdot  \frac{4 \cdot 4}{3 \cdot 5} \cdots  \frac{2n\cdot 2n}{(2n-1)(2n+1)} = I_{2n+1} \cdot \frac{1}{I_{2n}} \cdot \frac{2}{\pi}
$$
and with $I_n \ge I_{n+1} \ge I_{n+2}$ we have
$$
 1 \ge \frac{I_{n+1}}{I_n} \ge \frac{I_{n+2}}{I_n} = \frac{n+1}{n+2} \to 1 \quad (n \to \infty)
$$
so that $\frac{I_{n+1}}{I_n} \to 1$ for $n \to \infty$, which yields be manipulating limits Wallis product formula.
Now my question, in the above proof, that $I_n$ is decreasing was proved by using integral calculus, are there any more elementary ones, maybe just using the plain definition of the sequence?
 A: By applying the exponential generating function transformation for:
$$I_{n+2}-\frac{n+1}{n+2}I_n=0$$
where $\displaystyle P_0(n)=-\frac{n+1}{n+2}$ and $P_2(n)=1$, we get:
$$tY'+Y=tY'''+2Y''$$
which is hell of a equation to be solved:
$$Y(t) = \frac14 i \pi c_1 {\bf L_0}(t)+c_2 {\bf I_0}(t)+c_3 {\bf K_0}(t)$$
where $\bf i$, ${\bf L_n}(x)$, ${\bf I_n}(x)$ and ${\bf K_n}(x)$ are imaginary unit, modified Struve, modified Bessel function of first kind and modified Bessel function of the second kind respectively.
and expansion of the right side yields:
$$Y(t)=\frac14i\pi c_1\sum_{k=1}^{\infty}\frac{2t^{2k-1}}{((2k-1)!!\pi)^2}+c_2\sum_{k=0}^{\infty}\frac{t^{2k}}{4^k(k!)^2}+c_3\sum_{k=0}^{\infty}?$$
Unfortunately I couldn't write the expanded form of modified bessel function of second type.
Anyways using this we can show that:
$$I_n=\frac{(a_2(-1)^n+a_1)\Gamma\left(\frac{n+1}2\right)}{\sqrt{\pi}\;\Gamma\left(\frac n2+1\right)}$$
Using $I_0=\pi/2$ and $I_1=1$:
$$I_n=\frac{\sqrt{\pi}\;\Gamma\left(\frac{n+1}2\right)}{2\Gamma\left(\frac n2+1\right)}$$
Or in friendly notation:
$$I_n=\frac{\sqrt{\pi}(n+2)\left(\frac{n+1}2\right)!}{2(n+1)\left(\frac{n+2}2\right)!}$$
which surely is a decreasing function, see its graph:

A: I suppose the answer to your question is simply no. There is no way to prove something involving integrals without integrals(no pun intended:)).
