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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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You can throw away the first term since $$\dfrac{n\Gamma(n)}{ n^{n+1/2}} = \dfrac{n!}{n^{n+1/2}} \sim \dfrac{\sqrt{2 \pi n} (n/e)^n}{n^{n+1/2}} = \dfrac{\sqrt{2 \pi}}{e^n}$$ – user17762 Dec 31 '12 at 22:43
What is $\Gamma (n + 1, n)$? – gnometorule Dec 31 '12 at 22:46
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, $$e^n \Gamma(n+1,n) = \int_0^{\infty} (t+n)^n e^{-t}dt = \sum_{k=0}^n \int_0^{\infty} \dbinom{n}k t^k n^{n-k} e^{-t} dt = \sum_{k=0}^n \dbinom{n}k n^{n-k} \Gamma(k+1) = \sum_{k=0}^n n(n-1)(n-2) \cdots (n-k+1) n^{-k-1/2}$$ $$\dfrac{e^n \Gamma(n+1,n)}{n^{n+1/2}} = \sum_{k=0}^n \dbinom{n}k n^{-k-1/2} \Gamma(k+1) = \sum_{k=0}^n n(n-1)(n-2) \cdots (n-k+1) n^{-k-1/2}$$ – user17762 Dec 31 '12 at 22:51
up vote 2 down vote accepted

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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I think you mean $\sqrt{\pi/2}$ rather than $\pi/2$ in the line that starts "Hence, we get..." – user54551 Jan 2 '13 at 11:26
@lip1 Yes. Thanks. – user17762 Jan 2 '13 at 17:27

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