# Tagged Questions

Questions on the factorial function, $n!=n\times(n-1)\times\cdots\times1$. Consider using the tag (gamma-function) if dealing with noninteger arguments.

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

I've learned factorial. But today I saw a question which I don't know how to start with: $$\left(-\frac{1}{2}\right)!$$ Can anyone explain how to solve it? Thanks
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### How can $0!=1$ if the definition of factorial is $n!=n\times (n-1)!$ [duplicate]

Its a pretty basic question. If the definition of factorial is $n!= n\times(n-1)!$, then how can $0!=1$ since if we feed $0$ into the equation we get $0!=0\times (-1)!$? This comes after a ...
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### Is there a much simpler proof for Euler factorial formula?

Euler formula for factorial stated as follows Theorem [Euler]: For any non negative integers $a$ and $n$ such that $a\geq n$ $$n!=\sum_{k=0}^{n}(-1)^k\binom{n}{k}(a-k)^n$$ Proving this ...
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### How to calculate $\lim_{k \to \infty} \frac{(2k)!}{(2k+2)!}$

I need to calculate limit number 1, and I don't understand how to get out the factors. $$(1) \lim_{k \to \infty} \frac{(2k)!}{(2k+2)!}$$ $$(2) \lim_{k \to \infty} \frac{(k)!}{(k+1)!}$$ When I ...
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### function to approximate $x!$ without factorial

I am looking for a function $f(x)$ such that $f(x)\approx x!$, but (obviously) the function of x does not use factorial, eg a polynomial or exponential function. it does not have to be precise, just ...
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### Expressing $\sum_{k=1}^{n}\frac{1}{(k+2)k!}$ in terms of $n$.

How would I express $$\sum_{k=1}^{n}\frac{1}{(k+2)k!}$$ in terms of $n$? An attempt of mine is $$\sum_{k=1}^{n}\frac{1}{(k+2)k!} = \sum_{k=1}^{n}\frac{1}{(k+1)! + k!},$$ which is not useful for me. ...
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### Factorials…How do they do it?

So I've been recently arguing with my teacher about factorials. My teacher says that factorials can only be calculated for integers, because the definition of factorials is as follows: the product ...
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### Summation over a product of binomial coefficients

Question: I can't figure out why the following equality is true $\sum_\limits{k=a-b-c}^{d} (-1)^k \binom{d}{k}\binom{k+b+c}{a} = (-1)^d \binom{b+c}{a-d}$ How can this be shown? (In the book it just ...
186 views

### Sum of factorial fractions

Find the sum $$\sum\limits_{a=0}^{\infty}\sum\limits_{b=0}^{\infty}\sum\limits_{c=0}^{\infty}\frac{1}{(a+b+c)!}$$ I tried making something like a geometric series but couldn't. Then I couldn't think ...
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### Floor function of a factorial

Compute $$\left\lfloor \frac{1000!}{1!+2!+\cdots+999!} \right\rfloor.$$ How can I start with the problem? I thought of dividing by some number, but then I thought that some small numbers when added ...
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### Inverse question of trailing zeros [duplicate]

$(5n)!$ has $2014$ trailing zeros. What is $n$?
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### Growth of $n!!\dots !$

The asymptotic growth of the factorial function $n!$ is famously given by Stirling's formula as $$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$ Is there a similar formula for the iterated ...
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### Easy Double Sums Question: $\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{(m+n)!}$

How to calculate $\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \dfrac{1}{(m+n)!}$ ? I don't know how to approach it . Please help :) P.S.I am new to Double Sums and am not able to find any good sources ...
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### Does the inequality $n! > A \cdot B^{2n+1}$ hold for sufficiently large $n$?

Suppose $A,B >0$ are given constants. Is it possible to find a large enough $n \in \mathbb{N}$ such that $$n! > A \cdot B^{2n+1}?$$
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### Is there any way to extend the domain of this function through analytic continuation?

$\prod_{k=2}^x \log k=F(x)$ It looks a lot like the gamma function (a sort of logarithmic factorial), and I wonder if it can be similarly expressed as an integral or something. Any ideas?
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### Show $(2n+2)!\geq(n+2)(n+2)!$, $\forall n\in\mathbb{N}$.

It's the last step in a proof, and I just need to show that $$(2n+2)!\geq(n+2)(n+2)!$$ $\forall n\in\mathbb{N}$. I can't seem to do it though, any thoughts?
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### Calculate $\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!\cdot 2^n}$
Calculate the sum $$\displaystyle \sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!\cdot 2^n}$$ where $(2n-1)!!=1\cdot 3\cdots (2n-1)$, $(2n)!!=2\cdot 4 \cdots 2n$ Using Wolframalpha, the result is \$\sqrt{...
is there any way to simplify this expression or write it as a neat, concise formula? $$\frac{(2m)!}{2m!} - \frac{(x+y)!}{x!y!} \cdot \frac{ [2m-(x+y)]!}{ (m-x)!(m-y)!}$$ Thank you!