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Math Help - Proof by Induction

  1. #1
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    Proof by Induction

    I am to prove the following equation by induction:

    (1/2)+(1/6)+(1/12)+...+1/n(n+1)=n/(n+1)

    so far I have the base case= 1

    Assume this is true for n=k

    (1/2)+(1/6)+(1/12)+...+1/k(k+1)=k/(k+1)

    now I want to assume this is true for n=k+1

    (1/2)+(1/6)+(1/12)+...+1/(k+1)(k+2)=(k+1)/(k+2)

    This is where I am stuck...am I placing (k+1) in the correct places?
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  2. #2
    Senior Member roninpro's Avatar
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    You have a little problem here:

    now I want to assume this is true for n=k+1

    (1/2)+(1/6)+(1/12)+...+1/(k+1)(k+2)=(k+1)/(k+2)
    You shouldn't assume this. You need to show that this is true, based on the n=k case.

    Let's just look at the left hand side and try to work our way to the right.

    <br />
\frac{1}{2}+\frac{1}{6}+\ldots +\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}<br />

    Look at all of the terms except the last one. By the induction hypothesis, we can substitute them with \frac{k}{k+1} and receive

    <br />
\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}<br />

    Just clean this up and you will get what you want.
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  3. #3
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    How do we get the k/(k+1) on the left hand side of the equation via the induction hypothesis? Do I have to change the right hand side at all or is it just (k+1)/(k+2)?
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  4. #4
    Senior Member roninpro's Avatar
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    By the induction hypothesis, you have

    <br />
\frac{1}{2}+\frac{1}{6}+\ldots +\frac{1}{k(k+1)}=\frac{k}{k+1}<br />

    As I said earlier, you don't have a right hand side. You just start with
    <br />
\frac{1}{2}+\frac{1}{6}+\ldots +\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}<br />
and hope that you can work your way over to \frac{k+1}{k+2}.
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  5. #5
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    Quote Originally Posted by baker11108 View Post
    How do we get the k/(k+1) on the left hand side of the equation via the induction hypothesis? Do I have to change the right hand side at all or is it just (k+1)/(k+2)?
    Hi baker11108,

    P(k)

    \frac{1}{2}+\frac{1}{6}+......+\frac{1}{k(k+1)}=\f  rac{k}{k+1}

    P(k+1)

    \frac{1}{2}+\frac{1}{6}+....+\frac{1}{(k+1)(k+2)}=  \frac{k+1}{k+2}

    You want to try to prove that the formula being true for n=k causes the formula to be true for n=k+1

    Hence we write P(k+1) in terms of P(k) to see if there is a cause.

    Therefore if the sum of the first k terms really is \frac{k}{k+1}

    then the sum of k+1 terms is

    \frac{1}{2}+\frac{1}{6}+......+\frac{1}{k(k+1)}+\f  rac{1}{(k+1)(k+2)}

    will be \frac{k}{k+1}+\frac{1}{(k+1)(k+2)}

    which is

    \left(\frac{k+2}{k+2}\right)\frac{k}{k+1}+\frac{1}  {(k+1)(k+2)}

    =\frac{k^2+2k+1}{(k+1)(k+2)}=\frac{(k+1)^2}{(k+1)(  k+2)}=\frac{k+1}{k+2}

    Since this is what you get when you replace k with k+1 in the RHS
    it means that your hypothesis being true for some n=k
    causes the hypothesis to be true for the next n=k+1.

    Hence "true for n=1 causes it to be true for n=2 causes it to be true for n=3 causes......."
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  6. #6
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    Thanks very much. This helped a lot.
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