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Define

$$I_n=\int _0^{\frac{\pi}{2}} \sqrt{1+\sin^nx}\, dx$$

I have to show this sequence is convergent and find its limit. I proved it is decreasing: $\sin^{n+1} x \le \sin^n x \implies I_{n+1} \le I_n$. Also, it is bounded because:

$$0 \le \sin^n x \le 1 \implies \frac\pi{2} \le \int _0^{\frac{\pi}{2}} \sqrt{1+\sin^n x}\, dx\le \frac{\pi\sqrt{2}}{2}$$

so it is convergent. I'm stuck at finding the limit. I think it should $\frac{\pi}{2}$ but I'm not sure.

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  • $\begingroup$ You proved the sequence is bounded above, but you haven't shown it's convergent. $\endgroup$ – zhw. Feb 23 '20 at 20:21
  • $\begingroup$ @zhw. I showed it is bounded above and below. And also decreasing. So convergent, right? $\endgroup$ – user748957 Feb 23 '20 at 20:25
  • $\begingroup$ You are right. I missed the "decreasing" part. My bad. $\endgroup$ – zhw. Feb 23 '20 at 20:29
  • $\begingroup$ @user748957 If it is decreasing then you Have to show that the sequence is bounded below (which, in this case, is obvious since the lower bound is $0$) $\endgroup$ – Maximilian Janisch Feb 24 '20 at 7:50
  • $\begingroup$ @MaximilianJanisch, but I showed it is bounded below by $\frac{\pi}{2}$. Isn't that enough? $\endgroup$ – user748957 Feb 24 '20 at 8:31
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Let's define the following sequence:

$$J_n=\int _0^{\frac{\pi}{2}} \sin^nx\, dx$$

Since $\sin^n x \geq 0$, for $x \in [0,\frac{\pi}{2}]$, we have:

$$|\sqrt{1+\sin^n x}-1|=\frac{\sin^n x}{\sqrt{1+\sin^n x}+1} \leq \frac{\sin^n x}{2}$$

Therefore

$$1-\frac{1}{2}\sin^n x\leq \sqrt{1+\sin^n x}\leq 1+\frac{1}{2}\sin^n x$$

and integrating:

$$\frac{\pi}{2}-\frac{1}{2}J_n \leq I_n \leq \frac{\pi}{2}+\frac{1}{2}J_n\ \ \ \ \ \ \ \ (*)$$

Now, integrating by parts, we can deduce that:

$$J_n=\frac{n-1}{n}J_{n-2}\Rightarrow nJ_nJ_{n-1}=(n-1)J_{n-1}J_{n-2}$$

Therefore

$$nJ_nJ_{n-1}=(n-1)J_{n-1}J_{n-2}=(n-2)J_{n-2}J_{n-3}=...=J_1J_0=\frac{\pi}{2}$$

Clearly $J_n$ is convergent, and its limit must be $0$. Therefore, squeezing in $(*)$:

$$\lim_{n\to \infty}I_n = \frac{\pi}{2}$$

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    $\begingroup$ Your approach of proving $J_n\to 0$ is really cool. $\endgroup$ – Paramanand Singh Feb 24 '20 at 6:32
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Hint: Use Dominated convergence theorem.

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DCT is the way to go if you have that at your disposal. Here's an elementary solution: Let $0<b<\pi/2.$ Then the $n$th integral, let's call it $I_n,$ satisfies

$$\pi/2 = \int_0^{\pi/2}1\,dx < I_n < b\cdot \sqrt {1+\sin^n b} + \sqrt 2(\pi/2-b).$$

Taking limits, we get

$$\pi/2 \le \lim_{n\to \infty} I_n \le b\cdot 1 + \sqrt 2(\pi/2-b).$$

Now let $b\to \pi/2^-$ to see $\pi/2 \le \lim I_n \le \pi/2.$

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Take any $\delta \in (0,\pi/2).$ The sequence of integrable functions $f_n(x)=\sqrt {1+\sin^n x}$ converges uniformly to $1$ on $[0,\pi/2-\delta].$ So $\int_0^{\pi/2-\delta} f_n(x)dx$ converges to $\int_0^{\pi/2-\delta} 1\cdot dx=\pi/2 -\delta.$ So there exists $n_{\delta}\in \Bbb N$ such that $$n\ge n_{\delta}\implies -\delta<\int_0^{\pi/2-\delta}f_n(x)dx-(\pi/2-\delta)<\delta.$$ Meanwhile, for all $n$ we have $$0<\int_{\pi/2-\delta}^{\pi/2}f_n(x)dx<\int_{\pi/2-\delta}^{\pi/2}2\cdot dx=2\delta.$$ Therefore $$n\ge n_{\delta}\implies \pi/2-2\delta <\int_0^{\pi/2}f_n(x)dx <\pi/2+2\delta.$$

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