Tag Info

1

If the ring has a suitably nice measure defined on it, you might be able to adopt the Gamma or Pi functions as your notion of factorial.

1

$$n!=(n^2-n)(n-2)!=\prod\limits_{i=0,\\\,i\, \mathrm{even}}^{n-2}\left(n^2-(2i+1)n+i(i+1)\right)$$

2

As a way of working a problem like this in general, you can write the expression $(k + 1)^2$ at the left end of a line of your paper and the expression $(k + 1)!$ on the right end of the same line. But as you say, you cannot yet write $\le$ between them. (You might instead write a questoin mark.) Now underneath each of these expressions, start writing ...

5

There is no real need to use induction. Note that if $n\ge 3$ then $n!\ge (n)(n-1)(n-2)$. To prove that $n!\ge n^2$, it is enough to prove that $(n-1)(n-2)\ge n$, or equivalently that $n^2-4n+2\ge 0$. Since $n^2-4n+2=n(n-4)+2$, this is obviously true if $n\ge 4$.

5

The induction hypothesis is assumed. Induction hypothesis: Assume $k^2 \leq k!$. Now you use the inductive hypothesis to prove the inductive step: $$(k+1)^2 = k^2 + 2k + 1 \overset{?}{\leq} (k+1)! = (k+1)k!= kk! + k!$$ Can you see how to confirm that the inequality holds?

1

Hint: $$(k+1)!=(k+1).k!$$ ${}{}{}{}{}{}$

1

You're on the right track, but that only counts those containing exactly 6 ones. You want to count those containing at least 6 ones; meaning those containing 6, 7, or 8 ones. Use the same principle and add them together. $$\frac{8!}{2!6!}+ \frac{8!}{1!7!}+\frac{8!}{0!8!} = 37$$

4

Note that: $$(2n+2)! = (2n+2) \cdot (2n + 1) \cdot \underbrace{2n \cdot (2n - 1) \cdot (2n - 2) \dots \cdot 2 \cdot 1}_{=(2n)!}$$ Which means $$(2n+2)! = (2n+2) \cdot (2n+1) \cdot (2n)!$$ So when dividing $(2n+2)!$ by $(2n)!$ only those first two factors of $(2n+2)!$ remain (in this case in the denominator).

4

$\frac{(2n)(2n-1)(2n-2)...1}{(2n+2)(2n+1)(2n)(2n-1)(2n-2)...1} = \frac{1}{(2n+2)(2n+1)}$

3

Hint: One very useful property of factorials is that $$(N + 1)! = (N + 1) N! \implies \frac{(N + 1)!}{N!} = (N + 1)$$ Similarly, using the fact that $(N + 2)! = (N + 2)(N + 1) N!$ will help simplify the desired quotient.

1

The trick is to write $\dfrac{i}{(i+1)!} = \dfrac{1}{i!} - \dfrac{1}{(i+1)!}$ and use telescope. So the use of either geometric or arithmetic sequence formulas is not necessary.

5

$4N$ is not an upper limit. Every prime $p \leqslant N$ divides $N!$, with multiplicity $$m_p(N) = \sum_{k=1}^{\left\lfloor\frac{\log N}{\log p}\right\rfloor} \left\lfloor \frac{N}{p^k}\right\rfloor,$$ and no larger prime divides $N!$. We have $\frac{N}{p}-1 \leqslant m_p(N) < \frac{N}{p-1}$, and so $$\sum_{p\leqslant N} \left(\frac{N}{p}-1\right) ... 5 This does not completely answer your question but is helpful to know and the argument can be made completely rigorous. The highest power of a prime p dividing N! is$$\left\lfloor \dfrac{N}p \right\rfloor + \left\lfloor \dfrac{N}{p^2} \right\rfloor + \cdots \sim \dfrac{N}{p-1}$$and the number of primes less than N is \sim \dfrac{N}{\log(N)} and ... 2$$\frac{1}{n^n}\left(1 - \frac{1}{n}\right)^{n^2-n} \sim \frac{1}{n!} \iff \left(1 - \frac{1}{n}\right)^{n^2}\left(1 - \frac{1}{n}\right)^{-n} \sim \frac{n^n}{n!} \iff \left(1 - \frac{1}{n}\right)^{n^2} \sim \frac {e^{n-1}}{\sqrt{2\pi n}} \\ %\left(1 - \frac{1}{n}\right)^{n^2} %=\exp \left[ n^2\log \left(1-\frac 1n %\right) %\right] %=\exp \left[ n^2 ...

2

\begin{align} (k+1)! - 1 + (k+1)(k+1)! &= 1\cdot(k+1)! + (k+1)\cdot(k+1)! \\ &=\big((k+1)+1\big)\cdot(k+1)!-1 \\ &= (k+2)(k+1)! - 1 \\ &= (k+2)! - 1. \end{align}

1

Factor out $(k+1)!$ from the first equation. Then it should easily follow.

0

Use induction. For $n=1$ this is trivial. If the result holds for $n$, then $$(n+1)!-\sum_{k=1}^n k\cdot k! = (n+1)n!-n\cdot n!-\sum_{k=1}^{n-1} k\cdot k!=n!-\sum_{k=1}^{n-1} k\cdot k!=1$$ thus the result holds for all positive integers.

1

Hint: $k\cdot k! = (k+1)! - k!$

1

If you write it out, you will see that $82!+83!+84!=82\cdot 81\cdot 80\dots \cdot1+83\cdot 82\cdot 81\dots\cdot 1+84\cdot 83\cdot 82\cdot \dots\cdot 1$. You can factor $82!$ out, which leaves us with $82!(1+83+83\times 84)$. You can simplify this further. $$82!(1+83+83\cdot 84)$$ $$=82!(84+83\cdot 84)$$ $$=82!(1\cdot 84+83\cdot 84)$$ $$=82!(84\cdot 84)$$ ...

1

$82!+83!+84!=82!(1+83+83\cdot84)$

2

$$82! + 83! + 84!=82!(1+83+83\cdot84)$$ Now $\displaystyle1+n+n(n+1)=(n+1)^2$ Clearly, here $\displaystyle n=83$

1

We observe that $$n!> 2^n \quad (n\geq 4).$$ Indeed, for $n=4$ we have $4!=24>2^4=16$. Suppose that the inequality holds for $n=k$. Then $$(k+1)!=k!(k+1)>2^k(k+1)>2^k2=2^{k+1}.$$ Choose $n_0\in\mathbb{N}$ smallest such that $2^{n_0}>a$. Then $$n_0=\left \lfloor{\frac{\ln(a)}{\ln(2)}}\right \rfloor+1.$$ If $0\leq a<1$ then the least ...

3

Since $30!$ is divisible by all numbers from $2$ to $30$ obviously, $30!+2$ is divisible by $2$ $30!+3$ is divisible by $3$ $\vdots$ $30!+30$ is divisible by $30$ No primes.

2

Observe that $$n!+m$$ is divisible by $m$ for $2\le m\le n$ and integer $n\ge2$ So, we can have an arbitrarily large sequence of composite numbers for an arbitrary large value of integer $n$

0

$0! = 1$ usually is one of the "first principles".

0

The way, I think about it is via permutations. Suppose you have $n$ objects then to permute them you have $n!$ factorial ways. So, for example, if you had $3$ objects then to permute them you would do $3!$ which is $6$. Suppose, you had $0$ objects then how many times could you permute $0$ things? Once. Therefore, $0!$ should be $1$. This is just an ...

2

Also with Stirling's approximations, the comparison of numerical results :

0

Just extending what mookid gave, you can have an extremely good approximation using $$\frac{(2n - 2)!!}{(2n - 3)!!} \simeq \sqrt{\pi n } \left(1-\frac{3}{8 n}-\frac{7}{128 n^2}\right)$$

5


Top 50 recent answers are included