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Working on a harmonic analysis problem from the book, I came up with a difficulty of calculating the following sum. I figured how to solve original question, but it is a still interesting how to calculate this sum:

$$ \sum_{ \textit{ over odd values of } i=0}^{\min {n, k }}\frac{\Gamma(1 + m) \Gamma(1 - m + 2 n)}{(\Gamma(1 + i) Γ((1 - i) + m) \Gamma((1 - i) + n) \Gamma((1 + i) - m + n))}, $$ where $n \in N$, $k\geq 0$ - even.

metamorphy
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user4164
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    This is the same as $$ \sum_{\text{odd }i=0}^{\min(m,n)}\binom{m}i\binom{2n-m}{n-i} $$which is also $$ =\frac12\left(\sum_{i=0}^{\min(m,n)}\binom{m}i\binom{2n-m}{n-i}+\sum_{i=0}^{\min(m,n)}(-1)^{i+1}\binom{m}i\binom{2n-m}{n-i}\right) $$ The first summation is easy (Vandemonde's). The second at least has a nice answer when $m=n$, see this answer. When $m\neq n$, you can use a similar strategy, but it only lets you rewrite the summation as a different summation. – Mike Earnest Aug 24 '20 at 22:34
  • Thank you, Mike Earnest. I have gotten the Vandermonde's formula. My most confusion is the actual part when $m \neq n$. I am not sure on how to even represent it through other summations which would be helpful, i.e. which would be possible at least to estimate. – user4164 Aug 25 '20 at 03:25
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    Since that second sum is $-x^n^m(1+x)^{2n-m}$ using the coefficient-of notation, the other summations are obtained by rewriting it differently as a product of two known series, say $(1-x^2)^m(1+x)^{2(n-m)}$. Whether this is useful depends on what you need exactly (what are your values of $m$ and $n$, what kind of estimate do you need, etc.) – metamorphy Aug 25 '20 at 15:38

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