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 $1.25$.
Answer
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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