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Math Help - Simple induction help

  1. #1
    Member Jones's Avatar
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    Simple induction help

    Hi,

    I have a very simple induction problem that i'd like help solving.

    Prove that for all positive Integers n, that :
    \sum_{k=1}^{n} \frac{1}{k(k+1)} = \frac{n}{(n+1)}

    I know that the first step is to replace n with n+1 but then what?

    /Jones
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Jones View Post
    Hi,

    I have a very simple induction problem that i'd like help solving.

    Prove that for all positive Integers n, that :
    \sum_{k=1}^{n} \frac{1}{k(k+1)} = \frac{n}{(n+1)}

    I know that the first step is to replace n with n+1 but then what?

    /Jones
    The first step is to prove it is true for n=1.
    The second step is to state the inductive hypothesis, that is to say assuming that \sum_{k=1}^n \frac{1}{k(k+1)}=\frac{n}{n+1}

    The third step is to prove that it is true if you "replace" n by n+1, that is to say \sum_{k=1}^{n+1} \frac{1}{k(k+1)}=\frac{n+1}{n+2} :
    \sum_{k=1}^{n+1} \frac{1}{k(k+1)}=\left(\sum_{k=1}^n \frac{1}{k(k+1)}\right)+\frac{1}{(n+1)(n+2)}

    Now, use the inductive hypothesis to substitute \sum_{k=1}^n \frac{1}{k(k+1)}

    ----------------------------------------------
    Another way would have been to note that \frac{1}{k(k+1)}=\frac 1k-\frac 1{k+1} and use telescoping series (Telescoping series - Wikipedia, the free encyclopedia)
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  3. #3
    Member Jones's Avatar
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    Hmm, how do you mean?

    isn't it just a simple matter of:

    \sum_{k=1}^{n+1} \frac {1}{(k+1)(k+2)} = \frac{n+1}{n+2}
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  4. #4
    Super Member Showcase_22's Avatar
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    I'd do it like this:

    Proving it true for n=1:

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

    \frac {1}{1+1}=\frac{1}{2}


    True for n=1.

    That's the easy bit. I'll leave it up to you to do the rest.
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