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Proof by induction

  1. Feb 27, 2010 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    3. 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.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 27, 2010 #2
    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?
  4. Feb 27, 2010 #3
    I don't follow. Can you explain a bit or give an example?
  5. Feb 27, 2010 #4
    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?
  6. Feb 28, 2010 #5
    Suppose it's true for a certain n>= 2 then:

    [tex] \sum_{i=1}^{n+1} \frac{1}{i^2}} < 2-\frac{1}{n}+\frac{1}{(n+1)^2} < 2+\frac{1}{n+1}-\frac{1}{n}=2-\frac{1}{n(n+1)}<2-\frac{1}{n+1} [/tex]
  7. Feb 28, 2010 #6
    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?
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