# Thread: Inductive Questions, final algebra part

1. ## Inductive Questions, final algebra part

I need to show these are equal, but for the life of me I cant show it:
I know they involve a little trick to show but i just cant see it

1.
Prove that $\displaystyle 1/2! + 2 / 3!+ ... + n/(n+1)! = 1 - 1/(n+1)!$

$\displaystyle (k+1)/(k+2)!$ = $\displaystyle 1 - 1/(k+2)!$

2. Originally Posted by p00ndawg
I need to show these are equal, but for the life of me I cant show it:
I know they involve a little trick to show but i just cant see it
1.
Prove that $\displaystyle 1/2! + 2 / 3!+ ... + n/(n+1)! = 1 - 1/(n+1)!$
$\displaystyle (k+1)/(k+2)!$ = $\displaystyle 1 - 1/(k+2)!$
Suppose n=k is true:
$\displaystyle 1/2! + 2 / 3!+ ... + k/(k+1)! = 1 - 1/(k+1)!$

Prove: n=k+1

$\displaystyle 1/2! + 2 / 3!+ ... + k/(k+1)! + (k+1)/(k+2)! = 1 - 1/(k+2)!$

Left Side equals $\displaystyle = 1 - \frac{1}{(k+1)!} + \frac{(k+1)}{(k+2)!} =$

How does this compare to $\displaystyle 1-\frac{1}{(k+2)!}$? If they are equal, which is want we want to prove, then we need to show that:

$\displaystyle - \frac{1}{(k+1)!} + \frac{(k+1)}{(k+2)!} = -\frac{1}{(k+2)!}$

Equivalent to:
$\displaystyle \frac{1}{(k+1)!} - \frac{(k+1)}{(k+2)!} = \frac{1}{(k+2)!}$

Equivalent to:
$\displaystyle \frac{(k+2)-(k+1)}{(k+2)!} = \frac{1}{(k+2)!}$
which is clearly true.

I hope this helps!!

3. oh man thanks!

I didnt even THINK of subtracting the 1 at the beginning, and I thought I had tried everything!