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In a recent calculation I obtain a result involving the following expression depending on two integers $n,m\geq 0$: $$S(n,m):=\frac{(n+m+1)!}{n!m!}\sum_{l=0}^{n+m}\frac{1}{n+m-l+1}\sum_{\substack{j+k=l\\ 0\leq j\leq n\\0\leq k \leq m}}(-1)^{j}{n \choose j}{m \choose k}.$$ Numerical evidence suggests that one has $$ S(n,m)=\begin{cases} \sum_{k=0}^{m/2}{n+m+2 \choose 2k}, \quad & m\in \{0,2,4,\ldots\},\\ -\sum_{k=0}^{(m+1)/2-1}{n+m+2 \choose 2k+1}, & m\in \{1,3,5,\ldots\}.\end{cases} $$ Besides being much simpler, this formula has the great advantage that it makes obvious that $S(n,m)$ is an integer and non-zero.

How would I go about proving the claimed identity?

I guess it must be quite simple compared to many other binomial coefficient identities discussed here on MO. My hope is that experts in this business immediately see an "obvious" simplification step that I am not aware of.

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    $\begingroup$ The $L^+/L^-$, $\ell^+/\ell^-$ notation is very hard to parse (at least for me). I think it would be clearer to just use several different variables. $\endgroup$ May 28, 2020 at 13:17
  • $\begingroup$ @SamHopkins: Thank you, I just renamed the variables. $\endgroup$
    – B K
    May 28, 2020 at 13:29

1 Answer 1

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Taking into account that $\binom pq=0$ for nonnegative integers $p$ and $q$ such that $q>p$, write $$S(n,m)=\frac{(n+m+1)!}{n!m!}T(n,m),$$ where \begin{align*} T(n,m)&:=\sum_{l\ge0}\frac{1}{n+m-l+1}\sum_{\substack{j+k=l\\ j\ge0,k\ge0}}(-1)^{j}\binom nj\binom mk \\ &=\sum_{l\ge0}\int_0^1 dx\,x^{n+m-l}\sum_{\substack{j+k=l\\ j\ge0,k\ge0}}(-1)^{j}\binom nj\binom mk \\ &=\int_0^1 dx\,x^{n+m}\sum_{l\ge0}\sum_{\substack{j+k=l\\ j\ge0,k\ge0}}(-1)^{j}\binom nj\binom mk x^{-j} x^{-k} \\ &=\int_0^1 dx\,x^{n+m}\sum_{j\ge0}\binom nj(-x^{-1})^j\;\sum_{k\ge0}\binom mk x^{-k} \\ &=\int_0^1 dx\,x^{n+m}(1-x^{-1})^n(1+x^{-1})^m\ \\ &=\int_0^1 dx\,(x-1)^n(1+x)^m\ \\ &=\int_0^1 dx\,(x-1)^n\,\sum_{k=0}^m\binom mk x^k\ \\ &=(-1)^n\sum_{k=0}^m\binom mk\int_0^1 dx\,(1-x)^n x^k\ \\ &=(-1)^n\sum_{k=0}^m\frac{m!}{k!(m-k)!}\frac{k!n!}{(k+n+1)!}\ \\ &=(-1)^n \frac{m!n!}{(m+n+1)!} \sum_{k=0}^m\binom{m+n+1}{m-k}\ \\ &=(-1)^n \frac{m!n!}{(m+n+1)!} \sum_{j=0}^m\binom{m+n+1}j. \end{align*} So, \begin{equation} S(n,m)=(-1)^n\sum_{j=0}^m\binom{m+n+1}j, \end{equation} which is simpler than your desired expression (which seems to differ from the latter expression for $S(n,m)$ only by the sign).

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  • $\begingroup$ Thank you very much! I was blocked by the complication that one has to take into account the conditions $j\leq n$ and $k\leq m$ in the sum... of course it makes life easier to get rid of them as you did. $\endgroup$
    – B K
    May 28, 2020 at 15:11
  • $\begingroup$ @BK : I am glad this helped. $\endgroup$ May 28, 2020 at 16:19

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