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How can I find the upper bound of $\left\vert\frac{(c+1/2+\lambda)_{n}}{\lambda^{n}}\right\vert,\quad\text{where}\quad(c+1/2+\lambda)_{n}=\frac{\Gamma(c+1/2+\lambda+n)}{\Gamma(c+1/2+\lambda)}$ and $\lambda \to \infty$?

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    And the ratio decreases to its limit 1.2012-09-11

1 Answers 1

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$\left\vert\frac{(c+1/2+\lambda)_{n}}{\lambda^{n}}\right\vert =\left\vert\frac{(c+1/2+\lambda)}{\lambda}\right \vert\left\vert\frac{(c+1/2+\lambda -1)}{\lambda}\right\vert... \left\vert\frac{(c+1/2+\lambda -n+1)}{\lambda}\right\vert$

each term goes to 1 when $\lambda \to \infty$. so the limit is 1 as mentioned in the two comments above.