Interesting Limit (18)

Integrating by parts, we find $\frac{a_{n+2}}{a_n}=\frac{n+1}{n+4}$ so $\lim_{n \to \infty}\frac{a_{n+2}}{a_n}=1$ . Since the sequence $a_n$ is decreasing, we have $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=\boxed{1}$ as well.

• Substitute $x = sin\theta$
• so $dx = cos\theta d\theta$
• $\therefore a_{n} = \int_{0}^{\pi/2}sin^{n} \theta cos^{2}\theta d\theta$
• Then we can use walli's formula and canceling out the gamma function, $\int_{0}^{\frac{\pi }{2}}sin^{m}(x)cos^{n}(x) dx = \frac{\Gamma (\frac{m+1}{2}) \Gamma (\frac{n+1}{2})}{\Gamma (\frac{m+n +2}{2})}$

For large $n$ , $a_n \approx \int_0^1 x^n dx=\frac{1}{n+1}$ . So the ratio tends to $1$ .

A brute force: $\int_0^1 \sqrt{1-x^2} x^n \, dx$ for $n$ from 1 to 30: $\left\{\frac{1}{3},\frac{\pi }{16},\frac{2}{15},\frac{\pi }{32},\frac{8}{105},\frac{5 \pi }{256},\frac{16}{315},\frac{7 \pi }{512},\frac{128}{3465},\frac{21 \pi }{2048},\frac{256}{9009},\frac{33 \pi }{4096},\frac{1024}{45045},\frac{429 \pi }{65536},\frac{2048}{109395},\frac{715 \pi }{131072},\frac{32768}{2078505},\frac{2431 \pi }{524288},\frac{65536}{4849845},\frac{4199 \pi }{1048576},\frac{262144}{22309287}, \\ \frac{29393 \pi }{8388608},\frac{524288}{50702925},\frac{52003 \pi }{16777216},\frac{4194304}{456326325},\frac{185725 \pi }{67108864},\frac{8388608}{1017958725},\frac{334305 \pi }{134217728},\frac{33554432}{4508102925},\frac{9694845 \pi }{4294967296}\right\}$ .

A little of sequence matching gives the general formula of the sequence of integrals: $\frac{\sqrt{\pi } \left(\frac{n}{2}-\frac{1}{2}\right)!}{4 \left(\frac{n}{2}+1\right)!}$ .

The formula for the problem's quotient is $\frac{\frac{n}{2}! \frac{n+2}{2}!}{\frac{n-1}{2}! \frac{n+3}{2}!}$ giving a limit of 1.