Quote Originally Posted by kanoro View Post
So then I get summation as m=1 goes to k, 1/m+k - 8k+5 / ( 2k+2)(2k+1) . So how do I prove that they sum >0
The three fractions sum to a positive value (they only need sum to a non-negative value for the inductive proof).

Easiest is..

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

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

and as the fraction on the right has a bigger denominator, then it is less than the fraction on the left,
hence the subtraction yields a positive answer.

What this then means is....

if P(k) really is valid, P(k+1) will also be valid.
This is because if the sum of any number of k terms is >13/24,
then the sum of k+1 terms is greater again.
Hence, since the sum of 2 terms is >13/24, the sum of any (>2) number of terms is >13/24.