Limit with Integration

[man111]

[tex]\displaystyle \lim_{n\rightarrow \infty}\frac{\int^{\frac{\pi}{2}}_{0}(\sin x+\cos x)^{n+1}dx}{\int^{\frac{\pi}{2}}_{0}(\sin x+\cos x)^{n}dx}[/tex] for all [tex]n\in \mathbb{N}[/tex]

[Sup3rlum]

$\sqrt{2}$

[man111]

Yes Sup3rlum.

I have tried it

[tex]\displaystyle I_{n} = \int^{\frac{\pi}{2}}_{0}(\sin x+\cos x)^{n}dx = 2^{\frac{n}{2}}\int^{\frac{\pi}{2}}_{0}\cos\bigg(x+\frac{\pi}{4}\bigg)dx[/tex]

[tex]\displaystyle I_{n} = \int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\cos^{n}(t)dt[/tex]

So [tex]\displaystyle \sqrt{2}\lim_{n\rightarrow \infty}\frac{I_{n+1}}{I_{n}}[/tex]

plz help me how to show [tex]\displaystyle \lim_{n\rightarrow \infty}\frac{I_{n+1}}{I_{n}}=1[/tex]

Thanks

[Sup3rlum]

Okay so I think I finally figured out how to do it. I literally forced squeeze theorem until it worked.

You can cancel out twos from both integrals due to their evenness:

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}cos^nxdx=2\int_0^{\frac{\pi}{4}}cos^nxdx$

First observe that in the range $x \in (0, \frac{\pi}{4})$ the following inequality holds true for all $n \ge 1$

$cos^{n+1}(x)\le cos^n(x)\le cos^{n+1}\bigg(\frac{n-1}{n}x\bigg)$

Using the monotony of the integral of $cos^nx$ in that range, we can integrate everything:

$\int_0^{\frac{\pi}{4}}cos^{n+1}(x)dx\le \int_0^{\frac{\pi}{4}}cos^n(x)dx\le \int_0^{\frac{\pi}{4}}cos^{n+1}\bigg(\frac{n-1}{n}x\bigg)dx$

Divide everything by $\int_0^{\frac{\pi}{4}}cos^{n+1}(x)dx$:

$1 \le \frac{\int_0^{\frac{\pi}{4}}cos^n(x)dx}{\int_0^{\frac{\pi}{4}}cos^{n+1}(x)dx}\le \frac{\int_0^{\frac{\pi}{4}}cos^{n+1}\bigg(\frac{n-1}{n}x\bigg)dx}{\int_0^{\frac{\pi}{4}}cos^{n+1}(x)dx}$

The numerator integral on the rightmost side can be simplified by a substitution $u=\frac{n-1}{n}x$
$\Rightarrow dx=\frac{n}{n-1}du$

Using the linearity of the integral we can take this differential factor outside:

$1 \le \frac{\int_0^{\frac{\pi}{4}}cos^n(x)dx}{\int_0^{\frac{\pi}{4}}cos^{n+1}(x)dx}\le \frac{\frac{n}{n-1}\int_0^{\frac{\pi}{4}}cos^{n+1}(u)du}{\int_0^{\frac{\pi}{4}}cos^{n+1}(x)dx}$

As both integrals have the same numerical value, we can cancel them out and we will be left with:

$1 \le \frac{\int_0^{\frac{\pi}{4}}cos^n(x)dx}{\int_0^{\frac{\pi}{4}}cos^{n+1}(x)dx}\le \frac{n}{n-1}$

We can turn the inequality around and apply our limit towards infinity:

$\lim_{n \to \infty}1 \ge \frac{\int_0^{\frac{\pi}{4}}cos^{n+1}(x)dx}{\int_0^{\frac{\pi}{4}}cos^n(x)dx}\ge \frac{n-1}{n}$

As $n$ tends towards infinity $\frac{n-1}{n}$ becomes $1$, so we have:

$\Rightarrow \lim_{n \to \infty}1 \ge \frac{\int_0^{\frac{\pi}{4}}cos^{n+1}(x)dx}{\int_0^{\frac{\pi}{4}}cos^n(x)dx}\ge 1$

And by squeeze theorem we get that:

$\lim_{n \to \infty} \frac{\int_0^{\frac{\pi}{4}}cos^{n+1}(x)dx}{\int_0^{\frac{\pi}{4}}cos^n(x)dx} = 1$