# Can anyone solve this mathematical induction problem?

1. Oct 12, 2012

### ctothe

1. The problem statement, all variables and given/known data
Prove that for all n>1,
P(n) :$1 + 1/2 + 1/3 +....+1/n = k/m$

where k is an odd number an m is an even number.

2. Relevant equations

3. The attempt at a solution
1)Base Case n =2
P(2) = k/m
$3/2 = k/m$ which is true.

2) Inductive Step
Assume P(n) is true for some arbitrary n.

3) Prove P(n+1)
$P(n+1) = k/m$
$P(n) +1/(n+1) = k/m$
We know/assume that P(n) has an odd numerator and an even denominator. So,
$k/m + 1/(n+1) = k/m$
$(k(n+1) +m)/ (m+mn)$

So i divided the problem into two cases:

Case 1: n+1 is odd
if n+1 is odd then k(n+1) + m is odd and m +mn is even which is true.

Case 2: n+1 is even
here's where the problem is

2. Oct 14, 2012

### rude man

Let me rewrite your denominator mn + m = m(n+1).

Multiply m(n+1) by 2p.

Show that there exists a positive integer value for p such that the numerator must wind up odd whereas the denominator will always be even.

3. Oct 15, 2012

### CEL

k is odd. So, k(n+1) is odd, no matter if (n+1) is odd or even. Since m is even, k(n+1) + m is odd.

4. Oct 15, 2012

### rude man

Huh?

If k is odd and (n+1) is even, k(n+1) is even!

Odd x even = even always!