# Proof by induction

• Jul 5th 2009, 03:23 AM
LHS
Proof by induction
http://img30.imageshack.us/img30/2584/imageskr.jpg

I know roughly what has to be done here, but for some reason its not working out like the other questions, any help would be greatly appreciated!
• Jul 5th 2009, 03:32 AM
HallsofIvy
$\displaystyle T_n= \frac{1}{2}+ \frac{2}{3}+ \cdot\cdot\cdot+ \frac{n}{n+1}$
and you want to show that $\displaystyle T_n< n$.

Okay, if n= 1, $\displaystyle T_n= \frac{1}{2}< 1$.

Now suppose that, for some k, $\displaystyle T_k< k$. Now $\displaystyle T_{k+1}= T_k+ \frac{k+1}{k+2}< k+ \frac{k+1}{k+2}$. And you want to show that that is less than k+1.

Looks to me that you just need to show that $\displaystyle \frac{k+1}{k+2}< 1$!
• Jul 5th 2009, 03:43 AM
LHS
Ah I see, yes, it was at that stage I got confused.
I thought to prove it you would have to rearrange the RHS to show that the summation + the next term < k+1
Why do you need to show that (k+1)/(k+2) is less than one?
• Jul 6th 2009, 07:55 PM
AlephZero
Here is the "rearrangement" version you wanted:

$\displaystyle T_{k+1}=T_k+\frac{k+1}{k+2}<k+\frac{k+1}{k+2}$

$\displaystyle =\frac{k^2+3k+1}{k+2}=\frac{(k+2)(k+1)}{k+2}=k+1$

and you are done.

HallsofIvy's point was somewhat more elegant.
• Jul 6th 2009, 08:17 PM
malaygoel
Quote:

Originally Posted by AlephZero

$\displaystyle =\frac{k^2+3k+1}{k+2}=\frac{(k+2)(k+1)}{k+2}=k+1$

$\displaystyle k^2+3k+1=(k+2)(k+1)$

which is not true.
• Jul 6th 2009, 08:46 PM
AlephZero
Whoops, forgot to type a step. Here's the "rearrangment" proof:

$\displaystyle T_{k+1}=T_k+\frac{k+1}{k+2}<k+\frac{k+1}{k+2}$

$\displaystyle =\frac{k^2+3k+1}{k+2}<\frac{k^2+3k+2}{k+2}=\frac{( k+2)(k+1)}{k+2}=k+1.$

Thanks for pointing that out, malaygoel.
• Jul 6th 2009, 09:16 PM
simplependulum
$\displaystyle T_n = \frac{1}{2} + \frac{2}{3} + ... + \frac{n}{n+1}$

$\displaystyle = 1 - \frac{1}{2} + 1 - \frac{1}{3} + 1 - ... + 1 \frac{1}{n+1}$

$\displaystyle = 1+1+1+1 +... - \frac{1}{2} - \frac{1}{3} - ... - \frac{1}{n+1}$

$\displaystyle = n - \sum_{k=2}^{n+1} \frac{1}{k} < n$

for $\displaystyle n \geq 1$
• Jul 6th 2009, 10:13 PM
LHS
I think I see why that fraction has to be less than one. That right hand side should be k+1, or whenever you add the next term to the LHS you only add one to the RHS? So that fraction we gained from the LHS has to be less than 1 for the RHS to still be greater?
$\displaystyle \frac{k + 1}{k + 2} = \frac{k + 2 - 1}{k + 2}$
$\displaystyle = 1 - \frac{1}{k + 2} < 1$.