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 November 9th, 2017, 01:29 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 239 Thanks: 27 Limit with sequence How to find this limit? $\displaystyle \lim _{x\rightarrow 0}x^{-1}(1-\prod \limits _{j=1}^{n}\sqrt[j]{\cos jx}) =$ $\displaystyle \lim _{x\rightarrow 0} \frac{1-\cos x \cdot \sqrt{\cos 2x}\cdot \sqrt[3]{\cos 3x}\cdot ...\cdot \sqrt[n]{\cos nx}}{x}=\,?$ Last edited by skipjack; November 9th, 2017 at 02:34 AM.
 November 9th, 2017, 02:34 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,547 Thanks: 1754 What is the limit for n = 1? Can you find the limit for n = 2? Thanks from topsquark and idontknow
 November 9th, 2017, 04:31 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,405 Thanks: 2477 Math Focus: Mainly analysis and algebra I get a limit of zero using Taylor expansions. \begin{align*} \left. \begin{aligned} \cos{(x)} &= 1 - x^2 + \mathcal{O}(x^2) &\implies&& \cos{(nx)} &= 1 - (nx)^2 + \mathcal{O}(x^2) \\[8pt] &&&&\sqrt[n]{1 + x} &= 1 + \tfrac1nx + \mathcal{O}(x) \end{aligned} \right\} \implies \sqrt[n]{\cos{(nx)}} &= 1 - \tfrac1n(nx)^2 + \mathcal{O}(x^2) \\ &= 1 - nx^2 + \mathcal{O}(x^2) \end{align*} Thus the product \begin{align*} \prod_{n=1}^\infty \sqrt[n]{\cos{(nx)}} &= \prod_{n=1}^\infty \left(1 - nx^2 + \mathcal{O}(x^2) \right) \\ &= \left(1 - x^2 + \mathcal{O}(x^2) \right)\left(1 - 2x^2 + \mathcal{O}(x^2) \right)\left(1 - 3x^2 + \mathcal{O}(x^2) \right)\ldots \\ &= 1 + \mathcal{O}(x) & \text{(for some constant $a$)} \end{align*} by expanding the product term by term. Notice that, at this point we are assuming that the product has a finite value$^*$. We can motivate this by pointing out that for $x \lt \frac\pi2$, every term in the product is between $0$ and $1$, and thus the product must also be between $0$ and $1$. Now we can put this into the limit expression \begin{align*} \lim_{x \to 0} \frac{1 - \prod_{n=1}^\infty \sqrt[n]{\cos{(nt)}}}{x} &= \lim_{x \to 0} \frac{1 - \big(1 + \mathcal{O}(x)\big)}{x} \\ &= \lim_{x \to 0} \frac{ \mathcal{O}(x)}{x} \\ &= 0 \end{align*} https://www.desmos.com/calculator/b4rtdppucm $^*$ Note that the coefficient of $x^2$ is actually $\left(-\sum n\right)$ over the natural numbers. I assume that this leads to use of $\sum n = -\frac1{12}$. Thanks from topsquark and idontknow Last edited by v8archie; November 9th, 2017 at 05:19 AM.
 November 9th, 2017, 09:17 AM #4 Senior Member   Joined: Dec 2015 From: Earth Posts: 239 Thanks: 27 $\displaystyle \lim_{x \to 0} \frac{1 - \prod_{n=1}^\infty \sqrt[n]{\cos{(nt)}}}{x}=-\lim_{x \to 0} \frac{ e^{\ln \prod_{n=1}^\infty \sqrt[n]{\cos{(nt)}}}-1}{x}=-\lim_{x \to 0}\sum _{n=1}^{\infty}(nx)^{-1}\ln cos(nt)=0$
 November 9th, 2017, 09:50 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,405 Thanks: 2477 Math Focus: Mainly analysis and algebra I'm not convinced at all by that second equals sign.

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