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!
$\displaystyle T_n= \frac{1}{2}+ \frac{2}{3}+ \cdot\cdot\cdot+ \frac{n}{n+1}$
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 $\displaystyle \frac{k+1}{k+2}< 1$!
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.
$\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 $\displaystyle n \geq 1$
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?
Thanks guys for your help!