An integral limit!

$\displaystyle I_{n+1} = \int_{0}^{\pi/2}{{\left(\sin(x) + \cos(x)\right)}^{n+1} \ dx}$

$\displaystyle I_{n+1} = \int_{0}^{\pi/2}{{\left(\sin(x) + \cos(x)\right)}^n (\sin(x) + \cos(x)) \ dx}$

$\displaystyle I_{n+1} = \left. (-\cos(x) + \sin(x))(\sin(x) + \cos(x))^n \right]_{0}^{\pi/2} + n \int_{0}^{\pi/2}{(\cos(x) -\sin(x))^2 (\cos(x) + \sin(x))^{n-1} \ dx}$

$\displaystyle I_{n+1} = 2 + n \int_{0}^{\pi/2}{\left(2 - (\sin(x) + \cos(x))^2\right)(\sin(x) + \cos(x))^{n-1} \ dx}$

$\displaystyle I_{n+1} = 2 - n I_{n+1} + 2 n I_{n-1}$

$\displaystyle I_{n+1} = \frac{2}{n+1} + 2\frac{n-1}{n} I_{n-2}$

In other words $I_n = \frac{2}{n} + 2^2\frac{n-1}{n(n-2)} + 2^3 \frac{(n-1)(n-3)}{n(n-2)(n-4)} + \cdots$

Let $L = \lim_{n\rightarrow \infty} \frac{I_{n+1}}{I_n}$ ,

Then,

$\displaystyle L = \lim_{n\rightarrow \infty} \frac{2}{n I_n} + 2 (1) \frac{1}{L}$

$\displaystyle \lim_{n\rightarrow \infty} \frac{2}{n I_n} = \lim_{n\rightarrow \infty} \frac{2}{n\left(\frac{2}{n} + 2^2\frac{n-1}{n(n-2)} + 2^3 \frac{(n-1)(n-3)}{n(n-2)(n-4)} + \cdots\right)} = \lim_{n\rightarrow \infty} \frac{2}{\left(\frac{2}{1} + 2^2\frac{n-1}{(n-2)} + 2^3 \frac{(n-1)(n-3)}{(n-2)(n-4)} + \cdots\right)} = \lim_{n\rightarrow \infty} \frac{2}{\rightarrow \infty} = 0$

$\displaystyle L = 0 + \frac{2}{L}$

$\displaystyle \boxed{L = \sqrt{2}}$