Evaluating $\lim\limits_{n \to \infty }\sqrt[n]{ \frac{\left | \sin1 \right |}{1}\cdot\cdot\cdot\frac{\left | \sin n \right |}{n}}$ [closed]

Help me please with the limit:

$$\lim_{n \to \infty }\sqrt[n]{ \frac{\left | \sin1 \right |}{1}\cdot\frac{\left | \sin2 \right |}{2}\cdot\cdot\cdot\frac{\left | \sin n \right |}{n}}$$

Thanks!

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closed as off-topic by Daniel W. Farlow, ᴡᴏʀᴅs, 91500, 6005, Ofir SchnabelOct 27 '15 at 10:23

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The questions in the title and in the body are wildly different. – Did Nov 6 '12 at 7:20
@ did Oops..thanks – Lilly Nov 6 '12 at 7:53

$$\sqrt[n]{ \frac{\left | \sin1 \right |}{1}\cdot\frac{\left | \sin2 \right |}{2}\cdot\cdot\cdot\frac{\left | \sin n \right |}{n}} \leq \sqrt[n]{\dfrac1{n!}}$$ Now our good old Stirling's formula will do the job.

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+1 Nice solution. – copper.hat Nov 6 '12 at 7:16
There is no real need for Stirling. See math.stackexchange.com/questions/136626/…. – lhf Nov 6 '12 at 10:15

We have the following theorem, which is plain if you learned preliminary real analysis.

Theorem. If $a_n > 0$ and $\displaystyle \lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \ell$, then $\displaystyle \lim_{n\to\infty} \sqrt[n]{a_n} = \ell$.

Proof. We only consider the case $\ell > 0$. The case $\ell = 0$ easily follows by slightly modifying the proof of the case $\ell > 0$.

Now for any $\epsilon \in (0, \ell)$, there exists $N = N(\epsilon)$ such that whenever $n \geq N$, we have $$\ell - \epsilon < \frac{a_{n+1}}{a_n} < \ell + \epsilon.$$ Multiplying this inequality for $n$ replaced by $N, N+1, \cdots, n-1$ for $n > N$, we have $$(\ell - \epsilon)^{n-N} \leq \frac{a_n}{a_N} < (\ell + \epsilon)^{n-N}.$$ Taking $n$-th root and letting $n\to\infty$, we obtain $$\ell - \epsilon \leq \liminf_{n\to\infty} \sqrt[n]{a_n} \leq \limsup_{n\to\infty} \sqrt[n]{a_n} \leq \ell + \epsilon.$$ Since $\epsilon > 0$ is arbitrary, both limsup and liminf coincide with common value $\ell$, hence follows the claim.

Now you can apply this directly to conclude that the limit is zero.

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Terminology nit: The limit is zero, or the expression goes to zero. The limit doesn't go anywhere, it just is. – Harald Hanche-Olsen Nov 6 '12 at 9:20
@HaraldHanche-Olsen, I fixed it. – Sangchul Lee Nov 6 '12 at 10:09
A slightly more general results with almost identical proof: $\liminf \frac{a_{n+1}}{a_n}\leq \liminf \sqrt[n]{a_n} \le \limsup \sqrt[n]{a_n}\leq \limsup \frac{a_{n+1}}{a_n}$. See my comment here for links to several questions on MSE, where this appeared. – Martin Sleziak Nov 6 '12 at 10:52

$$\lim_{n\to\infty}\sqrt[n]{ \frac{\left | \sin1 \right |}{1}\cdot\frac{\left | \sin2 \right |}{2}\cdot\cdot\cdot\frac{\left | \sin n \right |}{n}} \leq \lim_{n\to\infty}{\dfrac1{\sqrt[n]{n!}}}=\lim_{n\to\infty}{\dfrac n{\sqrt[n]{n!}}}\cdot\frac{1}{n}=\lim_{n\to\infty} \frac{e}{n}=0.$$

Q.E.D. (we can safely apply d'Alembert criterion for $\lim_{n\to\infty}{\dfrac n{\sqrt[n]{n!}}})$

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With AM-GM the sequence is smaller than $\frac{H_n}{n}$ with Hn is équivalent ln (n) , then the limite is zéro.

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