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 $1.25$.
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 $1.25$.
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 $\sqrt{\dfrac{\pi}2} \approx 1.25$