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I am trying to solve summation I have stated in title, $$\sum_{0\leq n_1\leq n_2\leq\cdots\leq n_\beta\leq\gamma}{(-1)^{n_1+n_2+\cdots+n_\beta}} $$ where $\beta,\gamma\in\mathbb{N}$.

I have tried some methods, such as taking this summation as function of $\beta$ and $\gamma$, and make a recursive equation, $$f(\beta,\gamma)=f(\beta-1,0)-f(\beta-1,1)+\cdots=\sum_{k=0}^{\gamma}{(-1)^kf(\beta-1,k)}$$ Numerically I could just take this equation and make recursive function to calculate, but I want to know if there is any analytic solution to this summation.

Kenny Kim
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    Hint: $\sum\limits_{a\le n \le b} (-1)^n = \begin{cases}(-1)^a,& b \ge a \land b \equiv a \pmod 2\ 0,& \text{ otherwise }\end{cases}$ – achille hui May 25 '20 at 12:30
  • Here, as you can see from the problem there is several different ways to write $n_1+\cdots+n_\beta$, so with your hint, I could change my problem as $$\sum_{k=0}^{\beta\gamma}{(-1)^k\binom{\beta-1+k}{k}}$$ where $k=n_1+\cdots+n_\beta$ from the problem. Then I could not go further. – Kenny Kim May 26 '20 at 04:13

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There is a number of ways to proceed, but let's take your approach of $$f(\beta+1,\gamma)=\sum_{k=0}^{\gamma}(-1)^k f(\beta,k).$$ Define additionally $f(0,\gamma)=1$, so that the recurrence holds for all integers $\beta,\gamma\geqslant 0$.

Now, for $g(\beta,x)=\sum_{\gamma=0}^\infty f(\beta,\gamma)x^\gamma$, we get $$g(\beta+1,x)=\sum_{\gamma=0}^\infty x^\gamma\sum_{k=0}^\gamma(-1)^k f(\beta,k)=\sum_{k=0}^\infty(-1)^k f(\beta,k)\sum_{\gamma=k}^\infty x^\gamma=\frac{g(\beta,-x)}{1-x},$$ i.e. $g(2m,x)=(1+x)(1-x^2)^{-m-1}$ and $g(2m+1,x)=(1-x^2)^{-m-1}$ by induction. Thus, $$f(\beta,\gamma)=\begin{cases}\hfill 0,\hfill&\beta\text{ and }\gamma\text{ are odd}\\\frac{(\lfloor\beta/2\rfloor+\lfloor\gamma/2\rfloor)!}{\lfloor\beta/2\rfloor!\,\cdot\,\lfloor\gamma/2\rfloor!},&\hfill\text{otherwise}\hfill\end{cases}$$

metamorphy
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