# Thread: Simplifying Expression with Like Denominators and Factorials

1. ## Simplifying Expression with Like Denominators and Factorials

Hi,

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

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

2. ## Re: Simplifying Expression with Like Denominators and Factorials

Since (k+1)! = (k+1)k! = (k+1)k(k-1)!,

$\frac{1}{k!}-\frac{1}{(k+1)!} = \frac{1}{k!}\left(1-\frac{1}{k+1}\right)=\frac{1}{k!}\cdot\frac{k}{k+1 } =\frac{1}{(k-1)!(k+1)}$

3. ## Re: Simplifying Expression with Like Denominators and Factorials

Hi,

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

Is it wrong?

4. ## Re: Simplifying Expression with Like Denominators and Factorials

Originally Posted by SC313
Is it wrong?
No, both are correct. You should review the definition of factorial.

5. ## Re: Simplifying Expression with Like Denominators and Factorials

Could you please explain? I still am confused.

6. ## Re: Simplifying Expression with Like Denominators and Factorials

Have you reviewed and understood the definition of factorial?

7. ## Re: 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?

8. ## Re: Simplifying Expression with Like Denominators and Factorials

Yes, that what I meant. So, if you understand the definition, how can you not see that

$\frac{k}{(k+1)!}=\frac{k}{(k+1)k(k-1)\cdots1} =(\text{cancel }k)\frac{1}{(k+1)(k-1)\cdots1} =\frac{1}{(k+1)(k-1)!}$?

9. ## Re: Simplifying Expression with Like Denominators and Factorials

Originally Posted by SC313
the expression $\frac{1}{k!}-\frac{1}{(k+1)!}$.
$\frac{1}{k!}-\frac{1}{(k+1)!}=\frac{(k+1)}{(k+1)(k!)}-\frac{1}{(k+1)!}=\frac{k}{(k+1)!}$

10. ## Re: Simplifying Expression with Like Denominators and Factorials

Originally Posted by Plato
$\frac{1}{k!}-\frac{1}{(k+1)!}=\frac{(k+1)}{(k+1)(k!)}-\frac{1}{(k+1)!}=\frac{k}{(k+1)!}$
Do you not need like denominators? Should both denominators not be $(k+1)k!$?
It seems it would be $\frac{(k+1)}{(k+1)(k!)}-\frac{k!}{(k+1)!k!}$
I just really am frustrated and confused.

11. ## Re: Simplifying Expression with Like Denominators and Factorials

$(k+1)k! = (k+1)k(k-1)\cdots 2\cdot 1 = (k+1)!$

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

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

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

So, if you multiply $(k+1)!$ by $k!$, you don't get $(k+1)k!$.

12. ## Re: Simplifying Expression with Like Denominators and Factorials

Originally Posted by SC313
Hi,

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

To obtain like denominators, $\frac{1}{k!}-\frac{1}{(k+1)!}=\frac{(k+1)!}{k!(k+1)!}$
Notice that \displaystyle \begin{align*} \left( k + 1 \right) ! = \left( k + 1 \right) k! \end{align*}, so the lowest common denominator is \displaystyle \begin{align*} \left( k + 1 \right) k! \end{align*}, or just \displaystyle \begin{align*} \left( k + 1 \right) ! \end{align*}. So

\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*}