Dividing factorials

Question

I'm told that $\dfrac{(n+1)!}{(n+2)!}$ simplifies to $\dfrac{1}{n+2}$, but I dont understand how this works.

Could someone explain the theory of how to divide factorials like this?

Answer 1

HINT:

Example, set $n=5$:

$$\frac{(5+1)!}{(5+2)!}=\frac{6!}{7!}=\frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=\frac{1}{7}\cdot\frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{6\cdot5\cdot4\cdot3\cdot2\cdot1}=\frac{1}{7}\cdot1=\frac{1}{7}=\frac{1}{5+2}$$

Answer 2

$n!$ is pronounced $n$ factorial and mathematically is equal to the product $n!=n(n-1)(n-2)(n-3)(n-4) \ldots 5\cdot 4\cdot 3\cdot 2\cdot 1$. Factorials are defined for positive integers only and $0!$ is defined to be $1$ for our purpose only.

Quoting from this wikipedia link on factorials,

In mathematics, the factorial of a non-negative integer $n$, denoted by $n!$, is the product of all positive integers less than or equal to $n$. For example,

$$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120. \ $$ The value of $0!$ is $1$, according to the convention for an empty product.

The factorial function is formally defined by the product

$$n!=\prod_{k=1}^n k \!$$ or by the recurrence relation

$$ n! = \begin{cases} 1 & \text{if } n = 0, \\ (n-1)!\times n & \text{if } n > 0 \end{cases}$$ The factorial function can also be defined by using the power rule as

$$ n! = D^nx^n \ $$

Answer 3

For an algebraic proof:

$$ \require{cancel} \frac{(n+1)!}{(n+2)!}=\frac{(n+1)n(n-1)(n-2)(n-3)(n-4) \ldots 5\cdot 4\cdot 3\cdot 2\cdot 1}{(n+2)(n+1)n(n-1)(n-2)(n-3)(n-4) \ldots 5\cdot 4\cdot 3\cdot 2\cdot 1}=\frac{\cancel{(n+1)n(n-1)(n-2)(n-3)(n-4) \ldots 5\cdot 4\cdot 3\cdot 2\cdot 1}}{(n+2)\cancel{(n+1)n(n-1)(n-2)(n-3)(n-4) \ldots 5\cdot 4\cdot 3\cdot 2\cdot 1}}=\frac{1}{(n+2)}$$

Answer 4

\begin{align} (n+2)!&=(n+2)(n+1)(n)\dotsb(3)(2)(1)\\ (n+2)!&=(n+2)\cdot\big((n+1)(n)\dotsb(3)(2)(1)\big)\\ (n+2)!&=(n+2)\cdot(n+1)! \end{align} \begin{align} \frac{(n+1)!}{(n+2)!}&=\frac{\color{Red}{(n+1)!}}{(n+2)\cdot\color{Red}{(n+1)!}}\\ &=\frac1{n+2} \end{align}