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Given positive integers $k, m, n$ such that $1 \leq k \leq m \leq n$. Evaluate

$$ \sum^{n}_{i\mathop{=}0}\frac{1}{n+k+i}\cdot\frac{(m+n+i)!}{i!(n-i)!(m+i)!}$$

Any hints? I’m stuck on this one.

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Just a partial answer to collect some ideas.

Since

$\binom{m+n+i}{m+i}$

is the coefficient of $x^n$ in the Taylor series of $\frac{1}{(1-x)^{m+i+1}}$ around $x=0$, we have:

$$ S = [x^n]\left(\frac{1}{(1-x)^{m+1}}\cdot\sum_{i=0}^{n}\frac{1}{n+k+i}\binom{n}{i}\frac{1}{(1-x)^{i}}\right),\tag{1}$$

where $\frac{1}{n+k+i}=\int_{0}^{1}y^{n+k+i-1}dy$ leads to:

$$ S = [x^n]\left(\frac{1}{(1-x)^{m+1}}\cdot\int_{0}^{1}y^{n+k-1}\sum_{i=0}^{n}\binom{n}{i}\frac{y^{i}}{(1-x)^{i}}dy\right),$$

$$ S = [x^n]\left(\frac{1}{(1-x)^{m+1}}\cdot\int_{0}^{1}y^{n+k-1}\left(1+\frac{y}{1-x}\right)^n dy\right),$$

$$ S = [x^n]\left(\frac{(1-x)^{n+k}}{(1-x)^{m+1}}\cdot\int_{0}^{\frac{1}{1-x}}y^{n+k-1}(1+y)^n dy\right).\tag{2}$$

By setting $y=\frac{z}{1-z}$

$$ S = [x^n]\left(\frac{(1-x)^{n+k}}{(1-x)^{m+1}}\cdot\int_{0}^{\frac{1}{2-x}}\frac{z^{2n+k-1}}{(1-z)^{n+k+1}}dz\right)\tag{3}$$

we can see that the integral is just a value of the incomplete beta function, but I bet that other manipulations are more useful.

maxima’s Zeilberger tells me:

```
GosperSum(factorial(m + n + i)/(factorial(i) * factorial(n - i) * factorial(m + i) * (n + k + i)), i, 0, n);
NON_GOSPER_SUMABLE
```

so there isn’t a nice closed form.

I think you can find this

$$\sum^{n}_{i=0} \frac{(-1)^i}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!}$$

can see http://www.artofproblemsolving.com/Forum/viewtopic.php?p=239364&sid=2fbf367cb9fab8240df03e632a41085a#p239364

The sum is $$\frac{(m+n)!}{(n+k)n!m!}{}_3F_2(-n,n+k,m+n+1; m+1,n+1+k; -1)$$. Unfortunately there are not many closed form expressions known for these hypergeometric series at -1. A few like http://dx.doi.org/10.1016/S0377-0427(96)00154-9 may be able to map the format to $_3F_2(;;+1)$$ and be handled by Milgram’s table in http://arxiv.org/abs/1105.3126 .

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