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How can I compute the following limit?

$\lim_{n\rightarrow \infty} \frac{\frac{1}{n^n} \left(-\Gamma(n) n + e^n \Gamma(n+1,n)\right)}{\sqrt{n}}$

The answer appears to be about $1.25$.

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    And the second term is $\Gamma(n+1,n) = \int_n^{\infty} t^{n}e^{-t} dt = \int_0^{\infty} (t+n)^n e^{-(t+n)} dt$ Hence, $e^n \Gamma(n+1,n) = \int_0^{\infty} (t+n)^n e^{-t}dt = \sum_{k=0}^n \int_0^{\infty} \dbinom{n}k t^k n^{n-k} e^{-t} dt = \sum_{k=0}^n \dbinom{n}k n^{n-k} \Gamma(k+1) = \sum_{k=0}^n n(n-1)(n-2) \cdots (n-k+1) n^{-k-1/2}$ $\dfrac{e^n \Gamma(n+1,n)}{n^{n+1/2}} = \sum_{k=0}^n \dbinom{n}k n^{-k-1/2} \Gamma(k+1) = \sum_{k=0}^n n(n-1)(n-2) \cdots (n-k+1) n^{-k-1/2}$2012-12-31

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This is the same as the one worked out in this question. We have $\Gamma(n+1,n) = \dfrac{n!}{e^n} \left(\sum_{k=0}^n \dfrac{n^k}{k!}\right) \sim \dfrac{n!}2$ from this question. Hence, we get that $\dfrac{e^n \Gamma(n+1,n)}{n^{n+1/2}} \sim \dfrac{e^n n!}{2n^{n+1/2}} \sim \dfrac{e^n \sqrt{2 \pi} n^{n+1/2}}{2n^{n+1/2} e^n} = \sqrt{\dfrac{\pi}2}$ As I have in the comments, the first term can be thrown away. Hence, the limit is $\sqrt{\dfrac{\pi}2} \approx 1.25$

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    @lip1 Yes. Thanks.2013-01-02