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7 votes

Intuition behind the sum $\sum_{n=1}^{\infty} \frac {n-1}{n!}=1$?

This is just a telescoping sum: $$\sum_{n=1}^\infty \frac{n-1}{n!} = \sum_{n=1}^\infty \left(\frac{1}{(n-1)!} - \frac{1}{n!}\right) = \sum_{n=0}^\infty \frac{1}{n!} - \sum_{n=1}^\infty \frac{1}{n!} = \...
heropup's user avatar
  • 141k
5 votes
Accepted

Is this expression valid?

You are correct, the factorial is only defined for positive integers, so there are two possibilities here: Who made the twitter post intended $\lim_{x\to\infty}\frac{\Gamma(x+1)}{x^x}$, and wrote $x!$...
Zima's user avatar
  • 3,412
3 votes

Given nonzero $p(x)\in\mathbb Z[x]$. Are there infinitely many integers $n$ such that $p(n)\mid n!$ is satisfied?

Bober, Fretwell, Martin, and Wooley proved that when $F(x)\in\mathbb Z[x]$ is of the form $$ F(x)=\prod_{j=1}^\ell(a_jx^{k_j}-b_j),\tag1 $$ and $\varepsilon>0$ is arbitrary, there are infinitely ...
TravorLZH's user avatar
  • 7,193
2 votes

Find sum of factorials divisible by the largest possible prime squared

I used such approach: for given $n$ (currently, $n=32$), loop through prime numbers $p$ starting from certain value $p_0$ to, theoretically, $\sqrt{\sum_{k=1}^n k!}$; and for these $(n,p)$ construct 2 ...
Oleg567's user avatar
  • 17.4k
2 votes
Accepted

Determining the witnesses (constants $C_0$ and $k_0$) when showing $c^n \in O(n!)$ ($c > 1$)

Calculus is completely unnecessary here and so is Stirling's approximation. Let's do $c = 2$ as a warmup. How do we compare $2^n$ and $n!$? Well, observe that if $n \ge 2$ then $$\begin{align*} n! &...
Qiaochu Yuan's user avatar
1 vote

Is there an analog for factorials in division, and if so, what are its applications and properties?

As some of us have indicated in the comments to the OP, repeated division without grouping is ambiguous because division is not associative. That is, $$ (a \div b) \div c \not= a \div (b \div c) $$ ...
Brian Tung's user avatar
  • 34.5k

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