Induction style proof

  • Thread starter Bazman
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  • #1
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Main Question or Discussion Point

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:

Answers and Replies

  • #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:
  • #3
21
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thanks Murshid!

Its crystal clear now
 

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