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Explain this equality?

  1. Apr 7, 2014 #1
    1. The problem statement, all variables and given/known data
    ##\sum\limits_{n=1}^ \infty \frac{5^{n}}{n!} = \sum\limits_{n=0}^ \infty \frac{5^{n}}{n!} -1##


    2. Relevant equations

    3. The attempt at a solution

    How is this equality true? How does one get from the first to the second equation?
     
  2. jcsd
  3. Apr 7, 2014 #2

    micromass

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    [tex](a_1 + a_2 + a_3 .... ) = (a_0 + a_1 + a_2 + a_3 + ...) - a_0[/tex]
     
  4. Apr 7, 2014 #3

    phyzguy

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    What is the first term in [tex]\sum_0^\infty \frac{5^n}{n!}[/tex] ?
     
  5. Apr 7, 2014 #4
    Are you sure that the sum on the RHS is not between 2 and infinity?
     
  6. Apr 7, 2014 #5
    I'm certain.
     
  7. Apr 7, 2014 #6
    Thanks for the help. I get it now.
     
  8. Apr 7, 2014 #7
    Yes I get it too now! It's a 5 on the RHS not a 1, LOL!
     
  9. Apr 7, 2014 #8

    phyzguy

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    What do you mean? 5^0 = 1, 0! = 1.
     
  10. Apr 8, 2014 #9
    You need both to be true.
     
  11. Apr 8, 2014 #10

    micromass

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    What? The equality in the OP is entirely correct. It doesn't need to be a 5 or anything.
     
  12. Apr 8, 2014 #11

    Mark44

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    What's the first term in the series on the right side? Hint: It's NOT 5.
    ???
     
  13. Apr 8, 2014 #12

    Mark44

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    You're missing the point of this problem, which is strictly about manipulating the indexes of a summation. It's not about whether the series converges or not.
     
  14. Apr 8, 2014 #13
    Yes and you need that 5^0=1 and 0!=1.
     
  15. Apr 8, 2014 #14

    Mark44

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    These are true by definition. It wasn't clear from your previous comment about 5 on the right side, that you understood that both 5^0 and 0! were equal to 1.
     
  16. Apr 8, 2014 #15
    Thanks for checking, I'm fine with it now though the 0! had me stumped for a while!
     
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