### Autor Tema: limit

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27 Mayo, 2019, 03:54 pm
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#### jacks

• $$\Large \color{#c88359}\pi\,\pi\,\pi\,\pi$$
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For $a>0$. Then

$\displaystyle \lim_{n\rightarrow \infty}\frac{(\sin^{a}a+1)(\sin^{a}a+2)(\sin^{a}a+3)\cdots (\sin^{a}a+n)}{(a^{\sin a}+1)(a^{\sin a}+2)(a^{\sin a}+3)\cdots\cdot\cdot(a^{\sin a}+n)}$

28 Mayo, 2019, 12:13 am
Respuesta #1

#### Gustavo

• Mensajes: 1,843
• País: • Karma: +0/-0 ##### Re: limit
Hello,

The limit is not defined for all $a>0$ (for example, what if $\sin a<0$?). Whenever it is defined it seems that $0\le \sin^a a<a^{\sin a}$, but I didn't check this carefully. Assuming this...

Let $\alpha\ge 0$ and $\varepsilon>0$. The given limit is of the form $\displaystyle \lim_n \prod_{i=1}^n \dfrac{\alpha+i}{\alpha+\epsilon+i}$.

The convergence of the product is equivalent to the convergence of the sum

$\displaystyle \sum_i^n \log\left( \dfrac{\alpha+i}{\alpha+\varepsilon+i} \right) = \sum_i^n \log \left( 1-\dfrac{\varepsilon}{\alpha+\varepsilon+i} \right),$

which is divergent by the comparison test with $\displaystyle-\sum_i \dfrac{\varepsilon}{\alpha+\varepsilon+i},$ a harmonic series.

Moreover, $\displaystyle \sum_i^n \log\left( 1-\dfrac{\varepsilon}{\alpha+\varepsilon+i} \right) \to -\infty$   so   $\displaystyle\prod_{i=1}^n \dfrac{\alpha+i}{\alpha+\epsilon+i}\to 0.$