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Proving S2n-Sn is greater than a half in terms of the harmonic series

  1. Oct 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Let Sn = 1+[itex]\frac{1}{2}[/itex]+[itex]\frac{1}{3}[/itex]+[itex]\frac{1}{4}[/itex]...+[itex]\frac{1}{n}[/itex]. Show that |S2n-Sn|[itex]\geq[/itex] [itex]\frac{1}{2}[/itex]


    2. Relevant equations



    3. The attempt at a solution

    So I'm going to try and use induction

    Base case let n=1

    |S2n-Sn| = [itex]\frac{1}{2}[/itex]

    So true for base case

    Assume true for case that n=m

    So if n=m+1

    |S2m+2-Sm+1|= |S2m+2-S2m+S2m-Sm+1|

    Our induction assumption tells us that |S2m-S2m+2|[itex]\geq[/itex] [itex]\frac{1}{2}[/itex] and S2m+2-S2m = [itex]\frac{1}{2m+1}[/itex]+[itex]\frac{1}{2m+2}[/itex]> 0

    Therefore |S2m+2-Sm|= |S2m+2-S2m+S2m-S2m+2|> [itex]\frac{1}{2}[/itex]

    Have I used induction properly? Since I haven't used the base case at all the proof feels a little like I've made the assumption that it is true and then shown that it is true but at the same time I feel like I haven't broken any of the rules of induction. I'm I wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 24, 2012 #2

    LCKurtz

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    I don't quite follow that last step. But since ##\{S_n\}## is an increasing sequence you have$$
    |(S_{2m+2}-S_{2m}) + (S_{2m}-S_{m+1})|=(S_{2m+2}-S_{2m}) + (S_{2m}-S_{m+1})\ge \frac 1 2$$since the quantities in parentheses are positive.

    [Edit]: Ignore the highlighted paragraph. I wasn't completely awake. Look below:

    But this problem hardly needs induction. Write out ##S_{2n}-S_n##, look at how many terms there are, and underestimate each by the smallest one.
     
    Last edited: Oct 24, 2012
  4. Oct 25, 2012 #3
    So there are n terms between S2n and Sn the smallest of which is [itex]\frac{1}{2n}[/itex]

    It follows that |S2n - Sn|>n([itex]\frac{1}{2n}[/itex] ) =[itex]\frac{1}{2}[/itex]

    That is frustratingly simple. I feel silly but thanks a lot.
     
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