1. Not finding help here? Sign up for a free 30min 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!

This ought to be an easy sum

  1. Apr 30, 2006 #1

    benorin

    User Avatar
    Homework Helper

    I think I should know this already, but I don't; yet I need it, so would you please help me sum this...

    [tex]\sum_{n=k}^{\infty} \left(\begin{array}{c}n\\k\end{array}\right) \left( \frac{-z}{1-z}\right) ^{n} = (1-z)(-z)^{k}[/tex]
     
  2. jcsd
  3. Apr 30, 2006 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It's surely easier to look at:

    [tex]
    f(x) = \sum_{n=k}^{+\infty} \binom{n}{k} x^n
    [/tex]

    This is one of the basic combinatorial identities, but I must admit that I have forgotten it, and do not recall how to derive it. :frown:

    Hrm, maybe comparing f(x) to the integral of f(x)/x will help?

    Also, I notice that the k-th derivative of x^n is:

    [tex]
    \left(\frac{d}{dx}\right)^k \left( x^n \right) = \frac{k!}{x^k} \binom{n}{k} x^n
    [/tex]

    Oh, I think that does it. (Note that, in your sum, you may assume that n ranges over all nonnegative integers)
     
    Last edited: Apr 30, 2006
  4. Apr 30, 2006 #3

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    I was first thinking to look at derivatives of the left-hand side, and show that it too is a polynomial of degree k+1 with 0 being a zero of order k, but having no other zeroes where its defined (since its other zero would have to be at 1, but it's not defined there). However, the derivatives just got messy, so I couldn't make this work. The other idea was to express 1/(1-z) on the left as (1 + z + z² + ...). That turns the series on the left into a power series in z. It's easy to check that the resulting power series has 0's for coefficients of all powers of z before the kth power. It's easy to check also that the kth power of z has coefficient (-1)k and that the k+1th power of z has coefficient (-1)k+1.

    Somehow, you want to show that the rest have coefficient zero as well. Perhaps you can find a general expression for the jth coefficient in terms of previous coefficients, and solve by method of undetermined coefficients (inductively assuming something about the coefficients). Or maybe you can deal with the derivatives and show that the k+2th derivative is identically 0.
     
  5. Apr 30, 2006 #4

    benorin

    User Avatar
    Homework Helper

    Thanx Hurkyl!
     
  6. Apr 30, 2006 #5

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    Never mind, Hurkyl's way works better. Change variables x = -z/(1-z).

    [tex]\sum_{n=k}^{\infty} {{n}\choose{k}} \left( \frac{-z}{1-z}\right) ^n[/tex]

    [tex] = \sum_{n=0}^{\infty}{{n}\choose{k}} \left( \frac{-z}{1-z}\right) ^n[/tex]

    [tex] = \sum_{n=0}^{\infty}{{n}\choose{k}} x^n[/tex]

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

    [tex]= \frac{x^k}{k!}\sum_{n=0}^{\infty}\left(\frac{d}{dx}\right)^k(x^n) [/tex]

    [tex]= \frac{x^k}{k!}\left(\frac{d}{dx}\right)^k\sum_{n=0}^{\infty}x^n[/tex]

    [tex]= \frac{x^k}{k!}\left(\frac{d}{dx}\right)^k\left(\frac{1}{1-x}\right)[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: This ought to be an easy sum
  1. Easy series (Replies: 2)

  2. Lower sum/ upper sum (Replies: 1)

Loading...