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Math Help - Induction (involving inequality)

  1. #1
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    Induction (involving inequality)

    Hello,
    I am trying to understand the solution to a problem
    but I can't understand it. Could you explain it, please?
    Here is the problem:
    Prove by induction that

     1+ \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} \leq 2, n>0

    solution:
    Instead of proving  1+ \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} \leq 2, n>0
    we will prove something stronger:

     1+ \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} \leq 2 - \frac{1}{n} , n>0

    Basis: n=1: 1 \leq 1, holds

    Inductive step: The equation holds for n, so  1+ \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} \leq 2 - \frac{1}{n} , n>0.

    By adding to both sides \frac{1}{(n+1)^2} we have:

     1+ \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} + \frac{1}{(n+1)^2} \leq 2 - \frac{1}{n} +\frac{1}{(n+1)^2} <br />
 \leq 2 - \frac{1}{n+1} , so P(n+1) holds.

    Questions:

    1. How could someone think to prove
     1+ \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} \leq 2 - \frac{1}{n} (ie. adding - \frac{1}{n} )?

    2. Why should we add to both sides \frac{1}{(n+1)^2} ?

    3. In a similar problem, having to prove
    \frac{8}{1*3} + \frac{8}{5*7} + ... + \frac{8}{(4n+1)*(4n+3)} \leq 4

    we shall prove
    \frac{8}{1*3} + \frac{8}{5*7} + ... + \frac{8}{(4n+1)*(4n+3)} \leq 4 - \frac{1}{n+1}
    Why choose \frac{1}{n+1} and then add to both sides  \frac{8}{[(4(n+1)+1]*[4(n+1)+3]}?

    Thanks.
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  2. #2
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    Quote Originally Posted by drthea View Post
    1. How could someone think to prove
     1+ \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} \leq 2 - \frac{1}{n} (ie. adding - \frac{1}{n} )?
    The problem here is that if you tried to prove, 1+ \tfrac{1}{2^2}+ ... + \tfrac{1}{n^2} \leq 2 then by adding \tfrac{1}{(n+1)^2} to both sides (which becomes necessary because of induction) we get, 1+...+\tfrac{1}{(n+1)^2} \leq 2 + \tfrac{1}{(n+1)^2}, but we cannot continue any longer because 2 + \tfrac{1}{(n+1)^2} > 2 and induction breaks down here.

    2. Why should we add to both sides \frac{1}{(n+1)^2} ?
    That was already addressed on top. It is necessary because of induction.

    3. In a similar problem, having to prove
    \frac{8}{1*3} + \frac{8}{5*7} + ... + \frac{8}{(4n+1)*(4n+3)} \leq 4

    we shall prove
    \frac{8}{1*3} + \frac{8}{5*7} + ... + \frac{8}{(4n+1)*(4n+3)} \leq 4 - \frac{1}{n+1}
    Why choose \frac{1}{n+1} and then add to both sides  \frac{8}{[(4(n+1)+1]*[4(n+1)+3]}?
    Over here we might use \tfrac{1}{n+1} instead of \tfrac{1}{n} because we want to show this holds for n\geq 0. In order for this statement to make any sense we need to realize that \tfrac{1}{n} is a problem for then we have a zero denominator. There may be other reasons but this is the first one that I thought of.
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  3. #3
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    Thank you very much, your answer was very helpful
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