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Sum of Infinite Series via Unit Step Integration? Or Not?

  1. Oct 12, 2008 #1
    1. The problem statement, all variables and given/known data
    [sum of; n=1; to infinity] ((2y)^n)/(n(n!))

    2. Relevant equations

    3. The attempt at a solution
    If there were a way to find the improper integral of f(x) = ((2c)^x)/(x(x!)) from one to infinity using unit step integration (c is just a constant), then that would equal the sum, right? Well, I don't know how to do unit step integration, besides by tedious Riemann sums, but I can't do those to infinity. So, an explanation on unit step integration, or some other method of finding the sum of this series, is what I've been looking for for a while now.

  2. jcsd
  3. Oct 13, 2008 #2

    Gib Z

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    There is no such thing as "Unit step integration", and rightly so - that's *exactly* the same as taking summations. And if you were to set up the Riemann sums with unit widths, you would get the summation again.

    To sum the series, it would probably be best to let 2y= x, and find the derivative of the general term with respect to x. With some manipulation you should see that is just a constant times the general term of quite a well known series. I'm sure you can do the rest. Good luck and welcome to PF!

    EDIT: O damn it. It seems the result is only in terms of the Exponential integral. Hopefully you don't mind that? http://mathworld.wolfram.com/ExponentialIntegral.html. Either you use this method to see why in turns out that way, or you "recognize" the series.
    Last edited: Oct 13, 2008
  4. Oct 13, 2008 #3
    Shoot. I mean, of course thank you. It's just that I don't understand the exponential integral. I've looked at it and stuff, because I keep coming across it trying to solve this problem, but I guess I don't see how to apply it to my problem. It's the integral from -∞ to x of (e^t / t)dt? That.... seems to be how it's defined, but I don't think I yet understand what it actually is. What is this thing, what does it represent, how should it be applied here?
  5. Oct 13, 2008 #4


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    The exponential integral function Ei(x) is just a special function of x. You'll find it's relation with your series here http://en.wikipedia.org/wiki/Exponential_integral If you followed what Gib suggested you'll know that your series is represented by a function whose antiderivative is (exp(x)-1)/x and whose value at x=0 is 0.
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