# Homework Help: Proof Using General principle of math induction

1. Apr 29, 2010

### kolley

1. The problem statement, all variables and given/known data

prove that 1+1/4+1/9+...+1/n^2< or = 2-1/n for every positive integer n

2. Relevant equations

3. The attempt at a solution

proved it was correct for n=1, then replaced the n with k, changed it to k+1 to get:

1/(k+1)^2 < or = 2-1/(k+1)

don't know how to proceed
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 29, 2010

### Staff: Mentor

Note: <= means "less than or equal to." You don't need to write < or =.
You're not supposed to show that 1/(k+1)^2 <= 2-1/(k+1). You need to show that 1+1/4+1/9+...+1/(k + 1)^2 <= 2 - 1/(k + 1).

Your induction hypothesis is 1+1/4+1/9+...+1/k^2 <= 2 - 1/k. How can you get from this statement to the one you want to prove?

3. Apr 29, 2010

### kolley

Sorry, I left out part of mine. I had 49/36 +1/(k+1)^2 <= 2-1/(k+1)

since 1+1/4+1/9 is equal to 49/36, is this correct or am I still on the wrong track?

4. Apr 29, 2010

### jakncoke

Your part where u show the base case is correct but u can't just replace k with k+1. otherwise it would be a tautology not a proof. you have to show you can get it into the form where 1+1/4+1/9+...+1/k+1/(k+1) <= 2 - 1/(k+1)

Here is a simple proof: Show by induction that

$$1+2+3+...+n = \frac{n*(n+1)}{2} for n = 1 1*(1+1) = \frac{2}{2} = 1. so this is true for the base case.$$

now using rules of algebra if we add to one side, we add to the other, so

$$1+2+3+..+n+(n+1) = \frac{n*(n+1)}{2} + (n+1) = \frac{n^2+3n+2}{2} = \frac{(n+1)*(n+2)}{2}= \frac{(n+1)*((n+1)+1)}{2}$$

So by induction, we are done.

Last edited: Apr 29, 2010
5. Apr 29, 2010

### Staff: Mentor

You're still on the wrong track.

1+1/4+1/9+...+1/n^2 does not mean 1 + 1/4 + 1/9 + 1/n^2. The ellipsis - the three dots -- means "continuing in the same fashion." IOW, it means 1 + 1/4 + 1/9 + 1/16 + 1/25 + ... and so on, up to 1/n^2 for whatever value n happens to be.