find the limit of one sequence $\lim_{ n \to \infty } \cos(\frac{\pi}{2n + 1})\cos(\frac{2\pi}{2n + 1})...\cos(\frac{n\pi}{2n + 1}) $ I want to find the limit of this below Sequence.
the first sentence of this sentence n approaches infinity they approaches to 1 but not same for last sentence. How can we solve it.
$\lim_{ n \to \infty } \cos(\frac{\pi}{2n + 1})\cos(\frac{2\pi}{2n + 1})\cos(\frac{3\pi}{2n + 1})...\cos(\frac{n\pi}{2n + 1}) $
Is it possible to help me?
I'm sorry for bad English.
Thanks.
 A: First we prove
$$
\lim_{n\to \infty}\left(\cos\frac{\pi}{2n + 1}\cos\frac{2\pi}{2n + 1}...\cos\frac{n\pi}{2n + 1}\right)^{1/n}=\frac1{2}
$$
Take the log and we have
\begin{align}
\lim_{ n \to \infty }\ln \cos\frac{\pi}{2n + 1}\cos\frac{2\pi}{2n + 1}...\cos\frac{n\pi}{2n + 1}&=\lim_{n\to\infty}\sum_{k=1}^{n} \frac{1}{n}\ln \cos\left(k\cdot \frac{\pi}{2n+1} \right) 
\\
&=\int_{0}^{1} \ln \cos\frac{\pi t}{2} dt 
\\
&=\frac{2}{\pi}\int_0^{\pi/2}\ln{\cos{x}}\:dx
\\
&=\frac{2}{\pi}\int_0^{\pi/2}\ln{\sin{x}}\:dx
\end{align}
The last step is through $y=\pi/2-x$.
Let $I=\int_0^{\pi/2}\ln{\cos{x}}\:dx$. Then
\begin{align}
2I&=\int_0^{\pi/2}\ln{\cos{x}}\:dx+\int_0^{\pi/2}\ln{\sin{x}}\:dx
\\
&=\int_0^{\pi/2}\ln{\sin{x}\cos{x}}\:dx
\\
&=\int_0^{\pi/2}(\ln{\sin{2x}}-\ln{2})\:dx
\\
&=\int_0^{\pi/2}\ln{\sin{2x}}\:dx-\frac{\pi\ln{2}}{2}
\\
&=\frac1{2}\int_0^{\pi}\ln{\sin{x}}\:dx-\frac{\pi\ln{2}}{2}
\\
&=\frac1{2}\left(\int_0^{\pi/2}\ln{\sin{x}}\:dx+\int_{\pi/2}^{\pi}\ln{\sin{x}}\:dx\right)-\frac{\pi\ln{2}}{2}
\\
&=\frac1{2}\left(\int_0^{\pi/2}\ln{\sin{x}}\:dx+\int_{0}^{\pi/2}\ln{\cos{x}}\:dx\right)-\frac{\pi\ln{2}}{2}
\\
&=I-\frac{\pi\ln{2}}{2}
\end{align}
Hence $I=-\dfrac{\pi\ln{2}}{2}$. We conclude
\begin{align}
\lim_{n\to \infty}\left(\cos\frac{\pi}{2n + 1}\cos\frac{2\pi}{2n + 1}...\cos\frac{n\pi}{2n + 1}\right)^{1/n}&=\lim_{n\to \infty}e^{\frac1{n}\sum\limits_{k=1}^{n}\ln{\left(\cos\frac{k\pi}{2n+1}\right)}}
\\
&=e^{\frac{2}{\pi}I}=e^{-\ln{2}}
\\
&=\frac1{2}
\end{align}
Thus
$$
\lim_{n\to \infty}\cos\frac{\pi}{2n + 1}\cos\frac{2\pi}{2n + 1}...\cos\frac{n\pi}{2n + 1}=\lim_{n\to \infty}\frac1{2^n}=0
$$
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A: Each factor in the product lies between $0$ and $1.$ Thus the product lies between $0$ and the last factor, which is $\cos \pi(n/(2n+1)).$ As $n\to \infty,$ this factor $\to \cos (\pi/2) =0.$ Therefore the product converges to $0.$
