Autor Tema: limit

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27 Mayo, 2019, 03:54 pm
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jacks

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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

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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   \( \displaystyle\prod_{i=1}^n \dfrac{\alpha+i}{\alpha+\epsilon+i}\to 0. \)