5

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.

Community
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Proved Maroon Z
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  • Well the RHS looks more complicated than the other one. Why do you want to express it? BtW $\sum \frac{b}{\binom{q}{b}}>1+\frac{2}{q}$ – Phicar Jun 01 '17 at 04:26
  • I want to find expected value of random variable with distribution $P(b)=\frac{(n-q)!b!(q-b)!}{n!}$, so $\mathbf{E}b = {n \choose q} \sum_{b=0}^q\frac{b}{{q\choose b}}$ – Proved Maroon Z Jun 01 '17 at 04:36
  • I see that the random variable has a range [0,q], but what is n? – herb steinberg Dec 26 '17 at 22:09

2 Answers2

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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.

Tolaso
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1

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 $$\sum\limits_{k=1}^{n}k\binom{n}{k}^{-1}=\frac{n(n+1)}{2^{n+1}}\sum\limits_{k=0}^{n}\frac{2^k}{k+1}$$

user514787
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