# Inductive Questions, final algebra part

• October 15th 2009, 05:03 PM
p00ndawg
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 $1/2! + 2 / 3!+ ... + n/(n+1)! = 1 - 1/(n+1)!$

$(k+1)/(k+2)!$ = $1 - 1/(k+2)!$
• October 15th 2009, 06:53 PM
apcalculus
Quote:

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

Suppose n=k is true:
$1/2! + 2 / 3!+ ... + k/(k+1)! = 1 - 1/(k+1)!$

Prove: n=k+1

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

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

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

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

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

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

I hope this helps!!
• October 15th 2009, 07:02 PM
p00ndawg
oh man thanks!

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