Hi,

I am having difficulties simplifying the expression $\displaystyle \frac{1}{k!}-\frac{1}{(k+1)!} $.

To obtain like denominators, $\displaystyle \frac{1}{k!}-\frac{1}{(k+1)!}=\frac{(k+1)!}{k!(k+1)!} $

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- Nov 1st 2013, 04:33 PMSC313Simplifying Expression with Like Denominators and Factorials
Hi,

I am having difficulties simplifying the expression $\displaystyle \frac{1}{k!}-\frac{1}{(k+1)!} $.

To obtain like denominators, $\displaystyle \frac{1}{k!}-\frac{1}{(k+1)!}=\frac{(k+1)!}{k!(k+1)!} $ - Nov 1st 2013, 04:43 PMemakarovRe: Simplifying Expression with Like Denominators and Factorials
Since (k+1)! = (k+1)k! = (k+1)k(k-1)!,

$\displaystyle \frac{1}{k!}-\frac{1}{(k+1)!} = \frac{1}{k!}\left(1-\frac{1}{k+1}\right)=\frac{1}{k!}\cdot\frac{k}{k+1 } =\frac{1}{(k-1)!(k+1)}$ - Nov 1st 2013, 04:52 PMSC313Re: Simplifying Expression with Like Denominators and Factorials
Hi,

The answer according to Maple is $\displaystyle \frac{k}{(k+1)!} $.

Is it wrong? - Nov 1st 2013, 05:08 PMemakarovRe: Simplifying Expression with Like Denominators and Factorials
- Nov 1st 2013, 05:26 PMSC313Re: Simplifying Expression with Like Denominators and Factorials
Could you please explain? I still am confused.

- Nov 1st 2013, 05:28 PMemakarovRe: Simplifying Expression with Like Denominators and Factorials
Have you reviewed and understood the definition of factorial?

- Nov 1st 2013, 05:34 PMSC313Re: Simplifying Expression with Like Denominators and Factorials
Yes, n! is defined for a positive integer n as n!=n(n-1)...2*1.

Is that not what you mean? - Nov 1st 2013, 05:41 PMemakarovRe: Simplifying Expression with Like Denominators and Factorials
Yes, that what I meant. So, if you understand the definition, how can you not see that

$\displaystyle \frac{k}{(k+1)!}=\frac{k}{(k+1)k(k-1)\cdots1} =(\text{cancel }k)\frac{1}{(k+1)(k-1)\cdots1} =\frac{1}{(k+1)(k-1)!}$? - Nov 1st 2013, 05:46 PMPlatoRe: Simplifying Expression with Like Denominators and Factorials
- Nov 1st 2013, 07:33 PMSC313Re: Simplifying Expression with Like Denominators and Factorials
- Nov 1st 2013, 07:47 PMSlipEternalRe: Simplifying Expression with Like Denominators and Factorials
$\displaystyle (k+1)k! = (k+1)k(k-1)\cdots 2\cdot 1 = (k+1)!$

Example: $\displaystyle 5! = 5\cdot 4! = 5\cdot 4\cdot 3! = 5\cdot 4\cdot 3\cdot 2! = 5\cdot 4\cdot 3\cdot 2\cdot 1!$

Edit: In other words, they already had like denominators, but

$\displaystyle (k+1)!k! = [(k+1)k(k-1)\cdots 2\cdot 1][k(k-1)\cdots 2\cdot 1] = (k+1)k^2(k-1)^2\cdots 2^2\cdot 1^2$

So, if you multiply $\displaystyle (k+1)!$ by $\displaystyle k!$, you don't get $\displaystyle (k+1)k!$. - Nov 1st 2013, 08:47 PMProve ItRe: Simplifying Expression with Like Denominators and Factorials
Notice that $\displaystyle \displaystyle \begin{align*} \left( k + 1 \right) ! = \left( k + 1 \right) k! \end{align*}$, so the lowest common denominator is $\displaystyle \displaystyle \begin{align*} \left( k + 1 \right) k! \end{align*}$, or just $\displaystyle \displaystyle \begin{align*} \left( k + 1 \right) ! \end{align*}$. So

$\displaystyle \displaystyle \begin{align*} \frac{1}{k!} - \frac{1}{\left( k + 1 \right) !} &= \frac{k + 1}{\left( k + 1 \right) k!} - \frac{1}{ \left( k + 1 \right) ! } \\ &= \frac{k + 1}{ \left( k + 1 \right) ! } - \frac{1}{ \left( k + 1 \right) ! } \\ &= \frac{k}{ \left( k + 1 \right) ! } \end{align*}$