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.

  • $\begingroup$ As it is, it $\to 0$, since it is of the form $a_n = b_n \cos(\pi n / (2n+1))$ where $|a_n| < 1$ $\endgroup$ – reuns Nov 1 '16 at 4:01
  • $\begingroup$ Use $\prod_{r=1}^n \cos\frac{r\pi}{2n+1} = \frac{1}{2^n}$. $\endgroup$ – Jacky Chong Nov 1 '16 at 4:05

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 $$ \lim_{n\to \infty}\cos\frac{\pi}{2n + 1}\cos\frac{2\pi}{2n + 1}...\cos\frac{n\pi}{2n + 1}=\lim_{n\to \infty}\frac1{2^n}=0 $$ %%

  • $\begingroup$ Thanks for answer. Is it possible to prove that without Integral because my teacher said that we can solve without it too. we will learn it later. Is it possible? $\endgroup$ – Amin Nov 3 '16 at 11:37
  • $\begingroup$ Another proof is in @zhw post which is simpler. This is the proof that solves more general (harder) question that is normally asked: $\lim_{n\to \infty}\left(\cos\frac{\pi}{2n + 1}\cos\frac{2\pi}{2n + 1}...\cos\frac{n\pi}{2n + 1}\right)^{1/n}=\frac1{2}$ $\endgroup$ – Math Wizard Nov 3 '16 at 23:44

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 $0.$


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