1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    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
    Well, do you know what


    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
    The series


    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


    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
  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


    User Avatar
    Science Advisor

    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]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook