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November 9th, 2017, 02:29 AM   #1
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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 03:34 AM.
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November 9th, 2017, 03:34 AM   #2
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What is the limit for n = 1? Can you find the limit for n = 2?
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November 9th, 2017, 05:31 AM   #3
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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 $\sum n = -\frac1{12}$.
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Last edited by v8archie; November 9th, 2017 at 06:19 AM.
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November 9th, 2017, 10:17 AM   #4
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$\displaystyle \lim_{x \to 0} \frac{1 - \prod_{n=1}^\infty \sqrt[n]{\cos{(nt)}}}{x}=-\lim_{x \to 0} \frac{ e^{\ln \prod_{n=1}^\infty \sqrt[n]{\cos{(nt)}}}-1}{x}=-\lim_{x \to 0}\sum _{n=1}^{\infty}(nx)^{-1}\ln cos(nt)=0$
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November 9th, 2017, 10:50 AM   #5
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I'm not convinced at all by that second equals sign.
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