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$$\lim _{n\to \infty }\left(n\cdot \sqrt[n]{\sin\left(\frac{1}{1}\right)\cdot \sin\left(\frac{1}{2}\right)\cdot \cdot \cdot \sin\left(\frac{1}{n}\right)}\right)$$

I have noticed that the nth root is in fact the nth term in the series of geometric means of the series $ \sin\left(\frac{1}{n}\right)$.

Therefore i believe that the series behaves like $ \frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}}$ and therefore approaches 1. but i cant find a way of proving this.

I see that for each n we have $ \frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}} \le \left(n\cdot \sqrt[n]{\sin\left(\frac{1}{1}\right)\cdot \sin\left(\frac{1}{2}\right)\cdot \cdot \cdot \sin\left(\frac{1}{n}\right)}\right)$ and that this implies that the limit is greater than or equal to 1 but cant get much further than this.

This problem is meant to be solved without the notion of continuity and without certainly without taylor series expansion, lhospitals rule etc.. Any help would be appreciated. Thanks!

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  • $\begingroup$ maybe try a logarithm. $\endgroup$ – mathreadler Dec 11 '17 at 11:40
  • $\begingroup$ In reality, the limit is $e$. $\endgroup$ – Professor Vector Dec 11 '17 at 11:49
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Let $a_{n} =n^{n}\sin(1/1)\dots\sin(1/n)$ then $$\frac{a_{n+1}}{a_{n}}=\left(1+\frac{1}{n}\right)^{n}(n+1)\sin(1/(n+1))\to e$$ and therefore the given limit $\lim_{n\to\infty} a_{n} ^{1/n}$ is also $e$.

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$$\begin{align}\lim _{n\to \infty }\left(n\cdot \sqrt[n]{\sin\left(\frac{1}{1}\right)\cdot \sin\left(\frac{1}{2}\right)\cdot \cdot \cdot \sin\left(\frac{1}{n}\right)}\right)&=\lim _{n\to \infty }\left(n\cdot \sqrt[n]{\prod_{i=1}^n \sin\dfrac1i}\right) &\\=& \lim _{n\to \infty }\left(n\cdot \sqrt[n]{\prod_{i=1}^n \sin\dfrac1i}\right) &\\=& \lim _{n\to \infty }\left(\dfrac{n}{\sqrt[n]{n!}}\cdot \sqrt[n]{ \prod_{i=1}^n \dfrac{\sin (1/i)}{1/i}}\right)&\\=& \lim _{n\to \infty }\dfrac{n}{\sqrt[n]{n!}}\cdot\lim_{n\to \infty} \sqrt[n]{ \prod_{i=1}^n \dfrac{\sin (1/i)}{1/i}}&\\=& e\cdot\exp\left(\lim_{n\to\infty}\dfrac1n\sum^n_{i=1}\ln\left(\dfrac{\sin (1/i)}{1/i}\right)\right)&\\=& e\cdot\exp\left(0\right)=e \end{align}$$

I used Finding the limit of $\frac {n}{\sqrt[n]{n!}}$ and Convergence and value of infinite product $\prod^{\infty}_{n=1} n \sin \left( \frac{1}{n} \right)$?.

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Using the well-known inequality (see this question) $$x\left(1-\frac{x^2}6\right)\le\sin x\le x$$ for $x\ge0,$ we see $$\frac1{n!}\prod^n_{k=1}\left(1-\frac1{6k^2}\right)\le\prod^n_{k=1}\sin\left(\frac{1}{k}\right)\le\frac1{n!}.$$ The infinite product $$C=\prod^\infty_{k=1}\left(1-\frac1{6k^2}\right)$$ is convergent, i.e. $C>0$, so we obtain $$\sqrt[n]{C}\frac1{\sqrt[n]{n!}}\le\sqrt[n]{\prod^n_{k=1}\sin\left(\frac{1}{k}\right)}\le\frac1{\sqrt[n]{n!}}$$ and thus $$\sqrt[n]{C}\frac{n}{\sqrt[n]{n!}}\le n\,\sqrt[n]{\prod^n_{k=1}\sin\left(\frac{1}{k}\right)}\le\frac{n}{\sqrt[n]{n!}}.$$ Using Stirling's approximation $n!\sim\sqrt{2\pi\, n}\,n^n\,e^{-n}$, we see that $$\lim_{n\to\infty}n\,\sqrt[n]{\prod^n_{k=1}\sin\left(\frac{1}{k}\right)}=e.$$
BTW, from the $\sin x$ infinite product, we know that $$C=\frac{\sin\pi/\sqrt{6}}{\pi/\sqrt{6}},$$ but that isn't really important, here.

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