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Question: Let $f(x)=1-\sin x$ for $x\in \mathbb R$. Define $$a_n=\left(f\left(\frac{1}{n}\right)f\left(\frac{2}{n}\right)\cdots f\left(\frac{n}{n}\right)\right)^{1/n}.$$ Then $\lim_{n\to\infty }a_n=\dots?$ (exist finitely and $>0$ or infinitely)

My Attempt:

Given $a_n=(f(\frac{1}{n})f(\frac{2}{n})\dots f(\frac{n}{n}))^{1/n}$, taking $\log$ on both sides and thereafter using The Riemann Sum I am left with $$l=\exp\left({\int_{0}^{1}\log(1-\sin x) dx}\right),$$ where $l$ is the value of $\lim_{n\to\infty}a_n$ (existing finitely or infinitely). I am having trouble in dealing with this integral. Integration by parts is not helpful in this case. Is there anything special about this integral? I think it is closely linked with How to calculate $\int_0^\pi \ln(1+\sin x)\mathrm dx$ (not sure but may be! or might there be some other way to deal with this question.)

Thanks.

Thomas Andrews
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Aastha Choudhary
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    The Weierstrass substitution changes this to $$2\int_{0}^{\tan(1/2)}\frac{2\log(1-t)-\log(1+t^2)}{1+t^2},dt.$$ Not sure if that helps any. https://en.wikipedia.org/wiki/Tangent_half-angle_substitution – Thomas Andrews Oct 17 '22 at 15:38
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    You might be able to use the power series for $\log(1-z),$ since $|\sin(x)|\leq \sin(1)<1$ for $x\in[0,1],$ so the power series converges uniformly. But then you have to compute $\int_0^1\sin^n x,dx$ for each $n.$ Not impossible, but a bit of a pain. – Thomas Andrews Oct 17 '22 at 15:43
  • No issues, I will surely try. Thanks @ThomasAndrews – Aastha Choudhary Oct 17 '22 at 15:44
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    Is there a reason to expect a closed form? Like, is this a problem set problem asking for the value, or just the existence of a limit, which you've already shown? – Thomas Andrews Oct 17 '22 at 15:48

1 Answers1

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They don't ask you to compute that integral. All you need to show that it's finite. That's pretty easy since the function is monotonic on your interval, your first value is $0$, and the last value is $\ln(1-\sin 1)$. Since $1<\pi/2$, the sine function never reaches $1$, so the logarithm is always finite (and negative). Then $$0\ge \ln(1-\sin x)\ge \ln(1-\sin 1)$$so$$0\ge \int_0^1\ln(1-\sin x) dx\ge \ln(1-\sin 1)$$ Finally $$e^0=1\ge l\ge e^{\ln(1-\sin1)}=1-\sin 1\gt 0$$

Andrei
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