# Thread: stuck on two mathematical induction problems

1. ## stuck on two mathematical induction problems

Hello,

I have been trying to solve this problem for the last hour and haven't gotten anywhere. Got any ideas?

#1
(1 - 1/2)(1 - 1/2^2)...(1-1/n^2)=n+1/2n

Prove: (1-(1/(n+1)^2) + (n+1)/2n = (n+2)/(2n+2) for all integers >=2 (greater/equal)

#2
2^n < (n+2)!, for all integers n >= 0

2. ## Re: stuck on two mathematical induction problems

Originally Posted by dotman1989
Hello,

I have been trying to solve this problem for the last hour and haven't gotten anywhere. Got any ideas?

#1
(1 - 1/2)(1 - 1/2^2)...(1-1/n^2)=n+1/2n

Prove: (1-(1/(n+1)^2) + (n+1)/2n = (n+2)/(2n+2) for all integers >=2 (greater/equal)

#2
2^n < (n+2)!, for all integers n >= 0
#2

$\displaystyle 1) n=0 ; 2^0 < 2!$

$\displaystyle 2) \text{suppose that :} 2^n < (n+2)!$

$\displaystyle 3) \text{we have to prove :} 2^{n+1} < (n+3)!$

$\displaystyle 2 \cdot 2^n < 2 \cdot (n+2)! < (n+3)\cdot (n+2)!=(n+3)!$

$\displaystyle \text{hence:}$

$\displaystyle 2^{n+1} < (n+3)!$

3. ## Re: stuck on two mathematical induction problems

Originally Posted by dotman1989
Hello,

I have been trying to solve this problem for the last hour and haven't gotten anywhere. Got any ideas?

#1
(1 - 1/2)(1 - 1/2^2)...(1-1/n^2)=n+1/2n
Your real problem here, is that the "theorem", as stated, is not true. Specifically, if n= 2, the left side is (1- 1/2)(1- 1/4)= 1- 1/2- 1/4+ 1/8= 8/8- 4/8- 2/8+ 1/8= 9/8- 6/8= 3/8 while the right side is 2+ 1/4= 9/4, not 3/8.

Prove: (1-(1/(n+1)^2) + (n+1)/2n = (n+2)/(2n+2) for all integers >=2 (greater/equal)

#2
2^n < (n+2)!, for all integers n >= 0