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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!

  • 0
    @ did Oops..thanks2012-11-06

4 Answers 4

13

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

  • 3
    There is no real need for Stirling. See http://math.stackexchange.com/questions/136626/lim-limits-n-to-infty-sqrtnn-is-infinite.2012-11-06
10

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.

  • 1
    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](http://math.stackexchange.com/questions/116183/show-that-this-limit-is-equal-to-liminf-a-n1-n-for-positive-terms#comment372154_116183) for links to several questions on MSE, where this appeared.2012-11-06
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$\lim_{n\to\infty}\sqrt[n]{ \frac{\left | \sin1 \right |}{1}\cdot\frac{\left | \sin2 \right |}{2}\cdot\cdot\cdot\frac{\left | \sin n \right |}{n}} \leq \lim_{n\to\infty}{\dfrac1{\sqrt[n]{n!}}}=\lim_{n\to\infty}{\dfrac n{\sqrt[n]{n!}}}\cdot\frac{1}{n}=\lim_{n\to\infty} \frac{e}{n}=0.$

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.