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Series problem

  1. Jan 18, 2014 #1
    1. The problem statement, all variables and given/known data

    Prove the following result:
    \frac{1}{1!} + \frac{1+2}{2!} + \frac {1+2+2^2}{3!} ... = e2 - e

    2. Relevant equations



    3. The attempt at a solution

    Could someone please give me a hint on what to do . I tried writing out the maclaurin series for e^2 and e but don't know what to do with them.

    (also why won't the latex work for me?)
     
  2. jcsd
  3. Jan 18, 2014 #2

    Mentallic

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    The numerators are a geometric sum.
     
  4. Jan 18, 2014 #3

    Curious3141

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    Mentallic has already provided a hint - find an expression for the ##k##th term and put it in summation notation to make it clearer.

    And to format in LaTex, you have to enclose the text within tex tags like this (quote my post to see):

    [tex]\frac{1}{1!} + \frac{1+2}{2!} + \frac {1+2+2^2}{3!} ... = e^2 - e[/tex]

    or you can use the dollar sign shorthand notation like so:


    $$\frac{1}{1!} + \frac{1+2}{2!} + \frac {1+2+2^2}{3!} ... = e^2 - e$$
     
  5. Jan 18, 2014 #4
    Is the nth term, [1 + 2^(n-2) + 2^(n-1)]/n! ?
     
  6. Jan 18, 2014 #5

    Mark44

    Staff: Mentor

  7. Jan 18, 2014 #6

    Curious3141

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    Seems to work for n=3 but why don't you try other values of n?

    Mentallic already gave you the pattern for the numerator. How do you find the sum of that pattern?
     
  8. Jan 18, 2014 #7

    Mentallic

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    No, the nth term is

    [tex]\frac{1+2+2^2+2^3+...+2^{n-1}}{n!} = \frac{2^0+2^1+2^2+2^3+...+2^{n-1}}{n!}[/tex]
     
  9. Jan 19, 2014 #8

    HallsofIvy

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    [tex]= \frac{2^{n+1}- 1}{n!}[/tex]

    lionely did you not notice that the numerators are 1, 1+ 2= 3, 1+ 2+ 4= 7, 1+ 2+ 4+ 8= 15, 1+ 2+ 4+ 8+ 16= 31, etc., all one less than a power of two?
     
  10. Jan 20, 2014 #9

    Mentallic

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    Actually,

    [tex]\frac{2^n-1}{n!}[/tex]
     
  11. Jan 23, 2014 #10
    Would this be proving it , if I find the series for e^2 and e^1 and just add them? Because with the nth term I don't see how to show that is e^2 -e
     
  12. Jan 23, 2014 #11

    Mentallic

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    Can you apply the geometric sum formula for each numerator?
     
  13. Jan 23, 2014 #12
    What would the common ratio be , 2?
     
  14. Jan 23, 2014 #13

    Mentallic

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    Yes. You should brush up on geometric sums because by this point it's assumed that you can spot them and apply the rules correctly without hassle.
     
  15. Jan 23, 2014 #14
    I'm sorry but I'm still confused, I tried grouping the terms that are powers of two, but all the denominators are different. So how would I go about finding the sum to infinity of that?
     
  16. Jan 23, 2014 #15

    Mentallic

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    Show me what you have so far.
     
  17. Jan 23, 2014 #16
    so the series is (1/1!) + (1+2/2!) + (1+2+2^2/3!) .... right?

    so I tried grouping the 2's like this

    (1/1!) + (1/2!) +(3/3!) + 2(1/2! + 2/3! + 2^2/4! .....)
     
  18. Jan 23, 2014 #17

    Mentallic

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    No, you're going about this all wrong.

    We have

    [tex]\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+2^2}{2!}+\frac{1+2+2^2+2^3}{3!}+...+\frac{1+2+2^2+...+2^{n-1}}{n!}+...[/tex]

    Now we need to apply the geometric sum formula to each numerator. 1+2+22 can be simplified using the formula. 1+2+22+23 can be simplified. etc.

    What is 1+2+22 equivalent to using the formula?
     
  19. Jan 23, 2014 #18
    1/(1-2) ? also I just remembered just this only work if |r|<1?
     
  20. Jan 23, 2014 #19

    Mentallic

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    That obviously can't be right. Does

    [tex]1+2+2^2=\frac{1}{1-2}[/tex]

    ?

    You're thinking of infinite geometric summations, while we are dealing with finite sums. Just google geometric sum, read a little about it and it should all come back to you.
     
  21. Jan 23, 2014 #20
    Oh geometric sum, I kept thinking about summing to infinity. Um it would be (2^3-1)/2-1
     
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