Find $$ \lim_{n \rightarrow \infty} \left(n \int^{\frac{\pi}{4}}_0 (\cos(x)-\sin(x))^n \right)$$
I've managed to prove that the limit is in $(0,1]$ and I believe it is $1$ but I don't know how to prove it. Could you help me?
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asked Feb 21 '17 at 1:34
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1$\begingroup$ Are you sure it's in (0,1]? $n-n^2$ is negative, but the integral is positive. How can the limit of a sequence of negative terms be positive? $\endgroup$ – Clement C. Feb 21 '17 at 1:41
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$\begingroup$ I think that $n-n^2$ should be just $n$. $\endgroup$ – Sungjin Kim Feb 21 '17 at 2:50
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1$\begingroup$ You perhaps have a typo. The correct question is math.stackexchange.com/q/2121035/72031 $\endgroup$ – Paramanand Singh Feb 21 '17 at 4:18
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$\begingroup$ Oh, it was already answered with more precision in the link by Paramanand Singh. $\endgroup$ – Sungjin Kim Feb 21 '17 at 4:25
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$\begingroup$ @i707107: no worry you still get a +1 for your efforts from my side. $\endgroup$ – Paramanand Singh Feb 21 '17 at 4:30
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Using $\cos x = \sin(\pi/2 - x)$ and $\sin A - \sin B = 2\cos (\frac{A+B}2)\sin (\frac{A-B}2)$, we have $$ \cos x - \sin x = \sqrt 2 \sin(\frac{\pi}4 - x).$$ From this, we have $$ \int_0^{\frac{\pi}4} (\cos x - \sin x)^n dx = 2^{n/2} \int_0^{\frac{\pi}4} \sin^n (\frac{\pi}4 - x) dx= 2^{n/2}\int_0^{\frac{\pi}4} \sin^n x dx := 2^{n/2} I_n . $$ Let $J_n = n 2^{n/2} I_n$. Then by the reduction formula, we have $$ J_n = -1 + \frac{2(n-1)}{n-2}J_{n-2}. $$ Then $$ J_n-1 = \frac 2{n-2} + 2(1+\frac 1{n-2} ) (J_{n-2} -1) . $$ It is easy to check that $J_n$ is bounded, so $J_n-1$ is also bounded. Taking $\limsup$ on the right, we have for $\alpha = \limsup (J_n-1)$, $$ \alpha\geq 2\alpha. $$ Then take $\limsup$ on the left, we have $$ \alpha \leq 2\alpha. $$ Thus, $\alpha = 2\alpha = 0$. Similarly for $\liminf$, we obtain $\liminf (J_n-1) = 0$. Therefore $\lim J_n = 1$.
This shows that $$ \lim_{n\rightarrow\infty} n \int_0^{\frac{\pi}4} (\cos x - \sin x )^n dx = 1. $$
