Finite sum with inverse binomial I am looking for a closed-form for this summation:
$\sum_{b=1}^q\frac{b}{{q\choose b}}$
I looked at tables (Prudnikov-Brychkov book) of binomial sums but I couldn't find the result. WolpramAlpha answers:
$\sum_{b=1}^q\frac{b}{{q\choose b}}= -\frac{n _2F_1(1, n + 2; -q + n + 1, -1)}{q \choose n + 1} - \frac{_2F_1(1, n + 2; -q + n + 1, -1)}{q \choose n + 1} - \frac{_2F_1(2, n + 3; -q + n + 2, -1)}{q \choose n + 2} + \frac{_2F_1(2, 2; 1-q, -1)}{q}$
But when I was trying to calculate this expression (for example q=5) in MATLAB, last member of sum gives an error of invalid arguments, but direct computation is relatively easy returns 6.5.
Can anyone help me with simplifying this sum?
Thanks.
 A: I am not aware of a closed form but the following equality is very well known in literature:
$$\boxed{\sum_{k=0}^{n} k \binom{n}{k}^{-1} = \frac{1}{2^n} \left [ (n+1) \left ( 2^n-1 \right ) +\sum_{k=0}^{n-2} \frac{\left ( n-k \right )\left ( n-k-1 \right )2^{k-1}}{k+1} \right ]}$$
It relies upon a theorem of Mansour , see here along with other marvelous stuff. 
A: Using technique from robjohn's answer
$$2^{n+1}\binom{n+2}{2}^{-1}\sum\limits_{k=1}^{n}k\binom{n}{k}^{-1}=\sum\limits_{k=1}^{n}2^{k}\binom{k+2}{2}^{-1}\left(k+\sum\limits_{q=1}^{k}\frac{2^{q-k}}{q}\right)$$
After long time I came to similar question and it shows me how to get more simple result
$$2^{n+1}\binom{n+2}{2}^{-1}\sum\limits_{k=0}^{n}(k+p)\binom{n}{k}^{-1}=\sum\limits_{k=0}^{n}2^{k}\binom{k+2}{2}^{-1}\left(k+2p+(1-p)\sum\limits_{q=1}^{k}\frac{2^{q-k}}{q}\right)$$
then
$$2^{n+1}\binom{n+2}{2}^{-1}\sum\limits_{k=0}^{n}(k+1)\binom{n}{k}^{-1}=\sum\limits_{k=0}^{n}2^{k}\binom{k+2}{2}^{-1}(k+2)=\sum\limits_{k=0}^{n}\frac{2^{k+1}}{k+1}=\frac{2^{n+1}}{n+1}\sum\limits_{k=0}^{n}\binom{n}{k}^{-1}$$
so
$$\sum\limits_{k=0}^{n}(k+1)\binom{n}{k}^{-1}=\frac{n+2}{2}\sum\limits_{k=0}^{n}\binom{n}{k}^{-1}$$
and
$$\sum\limits_{k=1}^{n}k\binom{n}{k}^{-1}=\frac{n}{2}\sum\limits_{k=0}^{n}\binom{n}{k}^{-1}$$
which is also similar to
$$\sum\limits_{k=1}^{n}k\binom{n}{k}=\frac{n}{2}\sum\limits_{k=0}^{n}\binom{n}{k}$$
finally we get
$$\sum\limits_{k=1}^{n}k\binom{n}{k}^{-1}=\frac{n(n+1)}{2^{n+1}}\sum\limits_{k=0}^{n}\frac{2^k}{k+1}$$
