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)!} $
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)!} $
Yes, that what I meant. So, if you understand the definition, how can you not see that
$\displaystyle \frac{k}{(k+1)!}=\frac{k}{(k+1)k(k-1)\cdots1} =(\text{cancel }k)\frac{1}{(k+1)(k-1)\cdots1} =\frac{1}{(k+1)(k-1)!}$?
$\displaystyle (k+1)k! = (k+1)k(k-1)\cdots 2\cdot 1 = (k+1)!$
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 $\displaystyle (k+1)k!$.
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
$\displaystyle \displaystyle \begin{align*} \frac{1}{k!} - \frac{1}{\left( k + 1 \right) !} &= \frac{k + 1}{\left( k + 1 \right) k!} - \frac{1}{ \left( k + 1 \right) ! } \\ &= \frac{k + 1}{ \left( k + 1 \right) ! } - \frac{1}{ \left( k + 1 \right) ! } \\ &= \frac{k}{ \left( k + 1 \right) ! } \end{align*}$