Proving by Induction: P(n) for n>1

[yankees26an]

Homework Statement

Prove by induction that for an integer n where n>1 , http://img3.imageshack.us/img3/5642/prob1q.jpg

Homework Equations

The Attempt at a Solution

Prove P(2) is true
then prove P(x) = P(x+1) is true, then it's true for all x

That's all I really from proof by induction. It's just not very intuitive to me at all.

 

[bennyska]

it's been a little while since I've done this, so take this with a grain of salt. you'll probably want someone to verify it, or pretty it up.
so you've shown your base case, i.e. the statement holds for n=2. but you don't want to show P(x) = P(x+1). these are two different situations.
consider that the problem is saying 1/1^2 + 1/2^2 +...+1/n^2 < 2 - 1/n. you've shown it's true for n=2. now consider k, where k>2. what is this term? 1/k^2. what happens when you add 1/k^2 to both sides? is the statement still true?

 

[yankees26an]

what happens when you add 1/k^2 to both sides? is the statement still true?

I don't follow. Can you explain a bit or give an example?

 

[bennyska]

well, you've shown that 1/1^2 + 1/2^2 +...+1/n^2 < 2 - 1/n, when n = 2, right? so let's call 1/1^2 + 1/n^2 =a, and 2 - 1/n=b. so now a < b. so if we add 1/k^2, where k > 2, to both sides, we have a+1/k^2 < b+1/k^2. is this true? why?

 

[dirk_mec1]

Suppose it's true for a certain n>= 2 then:

\sum_{i=1}^{n+1} \frac{1}{i^2}} &lt; 2-\frac{1}{n}+\frac{1}{(n+1)^2} &lt; 2+\frac{1}{n+1}-\frac{1}{n}=2-\frac{1}{n(n+1)}&lt;2-\frac{1}{n+1}

 

[yankees26an]

Suppose it's true for a certain n>= 2 then:

\sum_{i=1}^{n+1} \frac{1}{i^2}} &lt; 2-\frac{1}{n}+\frac{1}{(n+1)^2} &lt; 2+\frac{1}{n+1}-\frac{1}{n}=2-\frac{1}{n(n+1)}&lt;2-\frac{1}{n+1}

Can you explain what you did?

Ok I'm still a bit confused

To show its true for n=2 I do

1/1^2+ 1/2^2 <(2 - 1/1) + (2- 1/2)

Is that how you prove it for n=2?