# Sum involving reciprocal of binomial coeffients

1. Mar 7, 2016

### amind

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. Mar 7, 2016

### Samy_A

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.

3. Mar 7, 2016

### amind

I got (n+1)*an+1-an-(n+1)
How do I convert an+1 to an ?

4. Mar 7, 2016

### Samy_A

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
5. Mar 7, 2016

### geoffrey159

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 $a_1,...,a_n$ ?

6. Mar 7, 2016

### amind

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)

7. Mar 7, 2016

### Samy_A

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 $a_n$.