Let's $a>0$ and $b>0$ be some, fixed real numbers. Let us consider sequence $S_m$, $m=0,1,\ldots$ defined as follow: $$ S_m = \sum_{n=0}^{\left\lfloor \frac{m-1}{2} \right\rfloor} \;\sum_{k=0}^{m - 2n - 1} \frac{(-1)^{n+k} \, a^{2n} \, b^{m - 2n - 1} \, \binom{n+k}{n}} {(2n + k + 1)! \cdot 2^{m - 2n - k - 1} \cdot (m - 2n - k - 1)!} $$ How to prove that $S_{2k} = 0$ for $k=0,1,\ldots$? Is it some nice formula for $S_{2k+1}$ for $k=0,1,\ldots$?
1 Answer
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Like a lot of questions about "sums of binomial coefficients", this follows easily from standard results on hypergeometric functions.
The sum over $k$ is a hypergeometric function with the variable specialized to $2$. Since it is a bit easier to find summations with variable $1/2$ in the literature, we reverse the order of summation, that is, we replace $k$ by $m-2n-1-k$. This gives $$S_m=\sum_{n=0}^{\lfloor(m-1)/2\rfloor}\binom{m-n-1}{n}\frac{(-1)^{m+n+1}a^{2n}b^{m-2n-1}}{m!}\,{}_2F_1\left(\begin{matrix}1+2n-m,-m\\ 1+n-m\end{matrix};\frac 12\right).$$ The relevant result is then Gauss's second summation theorem $${}_2F_1\left(\begin{matrix}a,b\\ (a+b+1)/2\end{matrix};\frac 12\right)=\frac{\Gamma(\frac 1 2)\Gamma(\frac{a+b+1}2)}{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}.$$ When $a=-n$ is a non-positive integer, this holds with the interpretation $${}_2F_1\left(\begin{matrix}-n,b\\ (b+1-n)/2\end{matrix};\frac 12\right)=\begin{cases}0, & n \text{ odd},\\ \displaystyle \frac{(\frac {1} 2)_{n/2}}{(\frac{1-b}2)_{n/2}}, & n \text{ even}. \end{cases}$$ In particular, $${}_2F_1\left(\begin{matrix}1+2n-m,-m\\ 1+n-m\end{matrix};\frac 12\right) =\begin{cases}0, & m \text{ even},\\ \displaystyle \frac{(\frac {1} 2)_{k-n}}{(k+1)_{k-n}}=\frac{k!(2k-2n)!}{(2k-n)!(k-n)! 2^{2k-2n}}, & m=2k+1. \end{cases} $$ It follows that $S_m=0$ for even $m$. For odd $m$, the sum over $n$ can be computed by the binomial theorem, and we obtain (as observed by Peter Taylor) $$S_{2k+1}=\frac 1{(2k+1)!}\sum_{n=0}^k\binom k n(-1)^na^{2n}(b/2)^{2k-2n}=\frac{(b^2/4-a^2)^k}{(2k+1)!}.$$
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$\begingroup$ I am confused: where are $a$ and $b$ from the original post in your computation? $\endgroup$Vladimir Dotsenko– Vladimir Dotsenko2025-04-24 13:26:36 +00:00Commented Apr 24 at 13:26
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1$\begingroup$ @VladimirDotsenko Already the inner sum (sum over $k$) vanishes for even $m$, and this sum is independent of $a$ and $b$. $\endgroup$Hjalmar Rosengren– Hjalmar Rosengren2025-04-24 13:36:13 +00:00Commented Apr 24 at 13:36
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3$\begingroup$ But when $m = 2k+1$ is odd, you should get $$S_{2k+1} = \frac{(\tfrac14b^2 - a^2)^k}{(2k+1)!}$$ $\endgroup$Peter Taylor– Peter Taylor2025-04-24 14:37:53 +00:00Commented Apr 24 at 14:37
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$\begingroup$ @PeterTaylor Thank you. That should follow from the explicit formula I indicate for the inner sum, combined with the binomial theorem. I will write some more details later. $\endgroup$Hjalmar Rosengren– Hjalmar Rosengren2025-04-24 15:06:15 +00:00Commented Apr 24 at 15:06
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$\begingroup$ @PeterTaylor Please see the new version where I corrected some typos and obtain the same result that you found $\endgroup$Hjalmar Rosengren– Hjalmar Rosengren2025-04-24 16:42:07 +00:00Commented Apr 24 at 16:42