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Finding sum of Power Series with Factorials

  1. May 13, 2009 #1
    1. The problem statement, all variables and given/known data

    I am confused about how to find a sum of a power series, especially when it contains factorials and I can't quite get it to look like a geometric series. Is it the same thing as finding a limit (and then I would follow the various tests for convergence of the different kinds of series)? Is it the same thing as finding an "exact value" (which I think means finding a function in terms of x)?

    2. Relevant equations

    I can't give the exact problems I am working with because it is a take home test, but I hope if I make up an example, this will work for my general questions about the concept:

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


    3. The attempt at a solution

    With one problem similar to above I am working with, when I did the ratio test it appears to converge to zero; does that mean the sum equals zero?
     
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  3. May 13, 2009 #2

    dx

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  4. May 13, 2009 #3
    I am having a similar problem myself in the above thread, but I do know that if you use the Ratio Test, and the outcome is less than one(i.e. zero) then the series converges, however this is NOT the sum. Past that i cant help. Sorry!
     
  5. May 13, 2009 #4
    Thanks, but, YIKES! That link was exactly the same question as on our test. I didn't look at your response to that question because I don't want to get into any trouble. I've looked into three calculus books and none of them explain this concept well enough. I was hoping to get some help with the general idea and then I can figure out the problem myself, but I don't want you to have to work double time to answer the same questions.
     
  6. May 13, 2009 #5

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    Hi PencilnPaper,

    If it's a take home test, I presume it's open book? If so, then I'm guessing it's ok for you to ask questions, as long as you don't ask any questions directly from your test.

    If you're at all unsure of whether you're allowed to seek help, then I think it's best if you don't.
     
  7. Apr 3, 2010 #6
    Think of the series as such:

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

    Now you know from the Maclaurin series expansion for [tex]e^x[/tex] that the following is true:

    [tex]\sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x[/tex]

    This is true because [tex]e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ... [/tex]

    Also note that [tex]e^x - 1 = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ... [/tex]

    Now the expansion of this original series is as follows: [tex] \sum_{n=0}^{\infty} \frac{x^n}{n!(n+1)} = 1 + \frac{x}{2*1!} + \frac{x^2}{3*2!} + \frac{x^3}{4*3!} + \frac{x^4}{5*4!} + ... [/tex]

    Subtract the first term (i.e. 1) from the expansion:

    [tex][\sum_{n=0}^{\infty} \frac{x^n}{n!(n+1)}] - 1 = \frac{x}{2*1!} + \frac{x^2}{3*2!} + \frac{x^3}{4*3!} + \frac{x^4}{5*4!} + ... [/tex]

    I'll put two particular results together:

    [tex][\sum_{n=1}^{\infty} \frac{x^n}{n!(n+1)}] = \frac{x}{2*1!} + \frac{x^2}{3*2!} + \frac{x^3}{4*3!} + \frac{x^4}{5*4!} + ... [/tex]

    [tex]e^x - 1 = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ... [/tex]

    Unfortunately, I don't know how to progress from here - I'm only in year 12, but maybe these ideas would help?

    Ulagatin
     
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