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Sum involving reciprocal of binomial coeffients

  1. Mar 7, 2016 #1
    1. The problem statement, all variables and given/known data
    If $$a_n = \sum_{r=0}^{n} \frac{1}{\binom{n}{r}}$$
    Find $$\sum_{r=0}^{n} \frac{r}{\binom{n}{r}}$$ in terms of an and n

    2. The attempt at a solution
    Let $$f(x) =\sum_{r=0}^{n} \frac{x^r}{\binom{n}{r}}$$
    Then, an = f(1).
    Observe that f'(1) is the required sum.
    I was thinking if I could find an expression for f(x) then I could obtain the required sum, but with no luck.
    Other than that I've tried expressing the binomial coefficients in terms of factorials, again in vain. So I am looking for some pointers that don't involve finding an by using Taylor series like I found on Google.
     
  2. jcsd
  3. Mar 7, 2016 #2

    Samy_A

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    Hint: notice that ##\displaystyle \frac{r}{\binom{n}{r}}=\frac{r r! (n-r)!}{n!}=\frac{(r+1) r! (n-r)!}{n!}-\frac{1}{\binom{n}{r}}=\frac{(r+1)! (n-r)!}{n!}-\frac{1}{\binom{n}{r}}=\frac{(n+1)(r+1)! (n+1-r-1)!}{(n+1)!}-\frac{1}{\binom{n}{r}}=\frac{n+1}{\binom{n+1}{r+1}}-\frac{1}{\binom{n}{r}}##

    Now take the sum for r=0 to r=n and see what it gives.
     
  4. Mar 7, 2016 #3
    I got (n+1)*an+1-an-(n+1)
    How do I convert an+1 to an ?
     
  5. Mar 7, 2016 #4

    Samy_A

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    I hoped that an expression including ##a_{n+1}## would be acceptable. If not, this is trickier than I thought.
    This paper gives a formula for ##a_n## (proof can be found here, the second proof is elementary), and from that you can derive ##a_{n+1}## in terms of ##a_n##. I doubt that this is how this exercise is intended to be solved. There must be an easier way ...
     
    Last edited: Mar 7, 2016
  6. Mar 7, 2016 #5
    And if you write ##f(r^\frac{1}{r}) = \sum_{r=0}^n \sum_{j=1}^r \frac{1}{C(n,r) }## and exchange the order of summation ? You will have ##f(r^\frac{1}{r})## function of ##a_1,...,a_n## ?
     
  7. Mar 7, 2016 #6
    Gotcha, the trick is writing r as n - (n-r) in the first step and then using the property C(n, r) = C (n, n-r)
     
  8. Mar 7, 2016 #7

    Samy_A

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    Nice.

    It gives ##\displaystyle \frac{r}{\binom{n}{r}}= \frac{n-(n-r)}{\binom{n}{r}}=\frac{n}{\binom{n}{r}}-\frac{(n-r)}{\binom{n}{n-r}}##, and then taking the sum for r=0 to r=n leads to the result.
    Indeed much simpler than actually computing ##a_n##.
     
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