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Serie Problem

  1. Aug 30, 2011 #1
    I know that the following serie converges to 2 (did in excel), still I would like to know how i can prove it step by step it.


    ∑ n/2^n
    n=1

    I tried (n+1)/(2^(n+1))/(n/2^n) still i'm finding 1/2, not the 2.

    Any thoughts?
     
  2. jcsd
  3. Aug 30, 2011 #2

    micromass

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    Well, do you know what

    [tex]\sum_{n=0}^{+\infty}{\frac{x^n}{2^n}}[/tex]

    converges to??

    And what if you take the derivative of that??
     
  4. Aug 30, 2011 #3
    Sorry, can you be more specific please?
     
  5. Aug 30, 2011 #4

    micromass

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    The series

    [tex]\sum_{n=0}^{+\infty}{\frac{x^n}{2^n}}[/tex]

    Can you find out what it converges to?? It's a nice geometric series...
     
  6. Aug 30, 2011 #5
    I did them in excel, found the following values.

    [tex]\sum_{n=0}^{+\infty}{\frac{n}{2^n}}[/tex] = 2/1

    [tex]\sum_{n=0}^{+\infty}{\frac{n}{3^n}}[/tex] = 3/4

    [tex]\sum_{n=0}^{+\infty}{\frac{n}{4^n}}[/tex] = 4/9

    [tex]\sum_{n=0}^{+\infty}{\frac{n}{5^n}}[/tex] = 5/16

    and on....

    So i figured out that i can find where any of these series (n/x^n) converge doing: x/(x-1)² where |x|<1 (it works to negative values too)

    I showed it to my professor, he says it's only a form of approximation, that i'm not proofing anything at all.

    The funny fact is that i can calculate

    [tex]\sum_{n=0}^{+\infty}{\frac{n}{-1024^n}}[/tex]

    And i know the value that serie will converge will be given by : (-1024/(-1024-1)²) = -9.74657941701368E-4

    Still i need a way to get to this... or a simple resolution of the previous given serie will help too.
     
  7. Aug 30, 2011 #6

    micromass

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  8. Aug 30, 2011 #7
    Sorry, spent like 1 hour in it and still can't find it. I'm tired today, thanks for the help anyway. I'm going to try tomorrow.
     
  9. Aug 31, 2011 #8

    HallsofIvy

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    Generally, people learn, in, say, a PreCalculus class, if not a basic algebra class, long before Calculus, that, if |r|< 1, then the geometric series
    [tex]\sum_{n= 0}^\infty ar^n= \frac{a}{1- r}[/tex]

    That is what micromass has been trying to remind you of. The series
    [tex]\sum \left(\frac{x}{2}\right)^n[/tex]
    is a geometric sequence with a= 1, r= x/2. As long as |x/2|< 1, i.e., as long as -2< x< 2, you can find the sum by the formula above.

    And, since
    [tex]\frac{d x^n}{dx}= nx^{n-1}[/tex]
    the derivative of that function has series
    [tex]\sum \frac{nx^n}{2^n}[/tex]
    which, taking x= 1, becomes
    [tex]\sum \frac{n}{2^n}[/tex]
     
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