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Sum of a power series in terms of x

  1. May 13, 2009 #1
    Hello, I need to find the sum(as a function of x) of the power series [tex]\Sigma^{\infty}_{n=0}\frac{(x+1)^n}{(n+2)!}[/tex]

    The hint i was given was compare it to the Taylor series expansion of ex.

    Im not sure even how to start this problem and any help is much appreciated.
     
    Last edited: May 13, 2009
  2. jcsd
  3. May 13, 2009 #2

    jbunniii

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    Do you know this series?

    [tex]\sum_{n=0}^{\infty} \frac{x^n}{n!}[/tex]
     
  4. May 13, 2009 #3
    yes, thats the expansion for ex. Its the (n+2)! in the original problem thats confusing me.
     
  5. May 13, 2009 #4

    dx

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    Factor out [tex] \frac{1}{(x + 1)^2} [/tex].
     
  6. May 13, 2009 #5
    ok i factored and came up with:

    [tex]

    \frac{1}{(x+1)^2} \sum_2^\infty \frac{(x+1)^n}{n!}

    [/tex]

    Do the indices matter? If not then it becomes

    [tex]\frac{e^{x+1}}{(x+1)^2}[/tex]
     
    Last edited: May 13, 2009
  7. May 13, 2009 #6

    dx

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    The indices (you should call them limits) do matter. What two terms do you have to add to

    [tex] \sum_2^\infty \frac{(x+1)^n}{n!} [/tex]

    to make it equal to ex+1?
     
  8. May 13, 2009 #7
    I would have to add the first two terms of the sum if the limits started at 0 which would be

    1 + (x+1)

    [tex]

    1 + (x+1) + \frac{e^{x+1}}{(x+1)^2}

    [/tex]?
     
  9. May 13, 2009 #8

    dx

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    Yes, the first two terms. But your answer is not correct.

    [tex] 1 + (1 + x) + \sum_2^\infty \frac{(x+1)^n}{n!} = e^{x+1} [/tex]

    This means

    [tex] \sum_2^\infty \frac{(x+1)^n}{n!} = e^{x+1} - 2 - x [/tex]

    Substitute this back in

    [tex] \frac{1}{(x+1)^2} \sum_2^\infty \frac{(x+1)^n}{n!} [/tex]
     
  10. May 13, 2009 #9
    [tex]

    \frac{e^{x+1} - x - 2}{(x+1)^2}

    [/tex]

    yes?
     
  11. May 13, 2009 #10

    dx

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    Correct.
     
  12. May 13, 2009 #11
    thank you for the help dx :!!)
     
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