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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.

user85798
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  • Maybe it have some connection to sums of binominals or multinomials? I mean $$ \sum^{n}{i\mathop{=}0}\frac{1}{n+k+i}\cdot\frac{(m+n+i)!}{i!(n-i)!(m+i)!} = \sum^{n}{i\mathop{=}0}\frac{1}{n+k+i}\cdot \binom{m+n+i}{i, n-i, m+i} = \ = \sum^{n}{i\mathop{=}0}\frac{1}{n+k+i}\cdot \binom{i}{i} \binom{n}{n-i} \binom{m+n+i}{m+i} = \sum^{n}{i\mathop{=}0}\frac{1}{n+k+i}\cdot \binom{n}{i} \binom{m+n+i}{m+i} $$ – Kusavil Feb 22 '14 at 21:39
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    The term is hypergeometric, so the techniques in A=B will tell you if there is a closed form. Computer algebra systems like maxima will do the heavy lifting for you. – vonbrand Feb 23 '14 at 22:41
  • @vonbrand Thanks for this cool book recommendation! – flonk Mar 06 '14 at 15:45
  • Do you require an exact value, or is an asymptotic expansion enough? Any special values (ranges) of the parameters of special interest? – vonbrand Mar 06 '14 at 16:04
  • Exact value is required. This is an olympiad problem. – user85798 Mar 06 '14 at 16:17
  • Probably not very helpful, but we can rewrite $i$-th element as a recursion: $$a_i=a_{i-1}\left(1-\frac{1}{n+k+i}\right)\frac{(m+n+i)(n-i+1)}{l(m+l)}$$ where $a_0=\frac{1}{n+k}$ and then the sum is $$\frac{(m+n)!}{n!m!}\sum_{i=0}^na_i.$$ – pisoir Mar 06 '14 at 18:33

4 Answers4

1

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.

Jack D'Aurizio
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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 .

R. J. Mathar
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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.

vonbrand
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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

math110
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