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Induction style proof

  1. Feb 26, 2007 #1
    Hi,

    I need to prove the following:

    [tex] 1+ \frac{ 1}{ 2!} + \frac{1 }{3!} +...+ \frac{ 1}{ n!} < 2 \lbrack 1 - ( \frac{ 1}{2 } )^n \rbrack [/tex]

    From trying various example I'm fairly sure the relation holds but I can't seem to prove it algebraically?

    Does the ineqaulity make a difference? Or can you behave pretty mcu as if it was an "=" ?

    I tried simply doing 2[1-(1/2)^n] + 1/(n+1)! to try to get to 2[1-(1/2)^n+1]
    but I can't seem to get very far?

    Can anyone shed any light?
     
    Last edited: Feb 26, 2007
  2. jcsd
  3. Feb 26, 2007 #2
    [tex]2 \left( 1 - \left( \frac{ 1}{2 } \right) ^n \right) = 1 + \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^{n-1}}[/tex]

    we also know that [tex]2^n < n![/tex]

    therefore,
    [tex]\frac{1}{(n+1)!} < \frac{1}{2^{n+1}}[/tex]

    so,
    [tex]2 \left( 1 - \left( \frac{ 1}{2 } \right) ^n \right) + \frac{1}{(n+1)!} < 2 \left( 1 - \left( \frac{ 1}{2 } \right) ^n \right) + \frac{1}{2^{n+1}}[/tex]

    [tex]2 \left( 1 - \left( \frac{ 1}{2 } \right) ^n \right) + \frac{1}{(n+1)!} < 2 \left( 1 - \left( \frac{ 1}{2 } \right) ^{n+1} \right) [/tex]
     
    Last edited: Feb 26, 2007
  4. Feb 26, 2007 #3
    thanks Murshid!

    Its crystal clear now
     
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