Wonder how to evaluate this factorial (−12)!\left(-\frac{1}{2}\right)!

I’ve learned factorial. But today I saw a question which I don’t know how to start with: (−12)! Can anyone explain how to solve it? Thanks Answer As suggested by Workaholic, I=(−12)!=Γ(12)=∫+∞0x−1/2e−xdx=2∫+∞0e−y2dy=∫Re−y2dy and by Fubini’s theorem: I2=∫R2e−(y2+z2)dydz=∫2π0∫+∞0ρe−ρ2dρdθ=π∫+∞02ρe−ρ2dρ=π so: (−12)!=√π. AttributionSource : Link , Question Author : Mathxx , Answer Author : Jack D’Aurizio

Does 625!625! have 156156 zeros at the end?

Someone wrote that 625!‘s last 156 digits are zeros because 125+25+5+1=156. If it’s true that 625! has 156 zeros at the end, how does “125+25+5+1=156” prove it? Answer When you factorize 625!, you are interested in the number of 10s you can divide it by until it is no longer possible to divide evenly. But … Read more

Find the limit lim\lim_{n\to\infty} \frac{(3n)! \, e^n}{(2n)!\,n^n\,8^n}

Find \lim_{n\to\infty} \frac{(3n)! \, e^n}{(2n)!\,n^n\,8^n} I tried by simplifying n! divided by n! = 1? What should I do next? I get then 3!e^n / 2!n^n8^n Answer HINT \frac{(3n)! e^n}{(2n)! n^n 8^n} = \left(\frac{e}{8n}\right)^n \prod_{k=1}^n (2n+k) = \prod_{k=1}^n \frac{2n+k}{8n/e} = \prod_{k=1}^n \left(\frac{e}{4} + \frac{ke}{8n}\right) AttributionSource : Link , Question Author : Andrej , Answer Author … Read more

Does series with factorials converge/diverge: $\sum\limits_{n=1}^\infty \frac{4^n n!n!}{(2n)!}$?

$$\sum_{n=1}^\infty {{4^n n!n!}\over{(2n)!}}$$ I tried the ratio test but got that the limit is equal to 1, this tells me nothing of whether the series diverges or converges. if I didn’t make any errors when doing the ratio test, it may diverge, but I need help proving that. Is there any other test I could … Read more

How to find lim\lim_{n\to \infty}{\frac{n^{n-1}}{n!}}

What method should I use for this limit? \lim_{n\to \infty}{\frac{n^{n-1}}{n!}} I tried ratio test but I ended with the ugly answer \lim_{n\to \infty}\frac{(n+1)^{n-1}}{n^{n-1}} which would go to 1? Which means we cannot use ratio test. I do not know how else I could find this limit. Answer Note that for n\ge 2 the top is … Read more

Finding lim\lim_{n \to \infty} \dfrac{n^n}{(2n)!}

Struggling to apply Squeeze THM to find this limit. Specifically, I need a sequence which is always larger than the one in the problem, but which can easily be derived from the middle sequence. Answer In the denominator (2n)!= (2n)(2n-1) \dots (n+1)\cdot n!. Each of the factors from (n+1) to (2n) are larger than n; … Read more

Does it make sense to multiply probabilities?

I got this interesting sum which seems to involve values of the derangement problem: ∞∑n=01(2n+2)(2n)!=e−1e=1−1e, where 1−1e is the probability that some man gets his own hat back [OEIS A068996]. Breaking down the intermediate expression to: 1e×(e−1), we have the probability that no man gets his own hat back [OEIS A068985], times the Engel Extension … Read more

Divergent sum of factorials

Is it possible to get an exact value of the sum (using divergent series summation methods) \sum_{n=0}^\infty~ \frac{(n+k)!}{n!} \quad? where k is a positive integer. The only other divergent sum of factorials I have seen is \sum_{n=0}^\infty(-1)^nn!. Does anyone know any useful techniques or references? Answer I don’t know how to check this, but here … Read more