# Show that $\lim_{m\rightarrow\infty}4^{-m}\sum\limits_{j=0}^{\lfloor m/2\rfloor}(-1)^j\binom{m}{j}\binom{3m-4j}{2m-4j}=1$ [closed]

I wish to show that the following sequence converge to $1$ as $m\rightarrow \infty$ $$\frac{\sum_{j=0}^{\lfloor m/2\rfloor}(-1)^j\binom{m}{j}\binom{3m-4j}{2m-4j}}{4^m}.$$ Any idea of how to do this?

Consider the definition of the Backward Finite Difference $$\begin{gathered} \nabla _{\,x} f(x) = f(x) - f(x - 1) \hfill \\ \nabla _{\,x} ^2 f(x) = \nabla _{\,x} \left( {\nabla _{\,x} f(x)} \right) = f(x) - 2f(x - 1) + f(x - 2) \hfill \\ \quad \quad \vdots \hfill \\ \nabla _{\,x} ^n f(x) = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( { - 1} \right)^k \left( \begin{gathered} n \hfill \\ k \hfill \\ \end{gathered} \right)f(x - k)} \hfill \\ \end{gathered}$$ It is not difficult to demonstrate that in the case of having a polynomial of degree $m$ we obtain $$\nabla _{\,x} ^m p_{\,m} (x) = \nabla _{\,x} ^m (a_{\,m} x^{\,m} + a_{\,m - 1} x^{\,m - 1} + \; \cdots \; + a_{\,0}) = m!a_{\,m}$$ Now, in your case, because of the limited excursion of the sum index, we have in general that $$\sum\limits_{0\, \leqslant \,j\,\left( { \leqslant \,\left\lfloor {m/2} \right\rfloor \, \leqslant \,m} \right)} {\left( { - 1} \right)^j \left( \begin{gathered} m \\ j \\ \end{gathered} \right)\left( \begin{gathered} 3m - 4j \\ 2m - 4j \\ \end{gathered} \right)} \quad \left| {\;0 \leqslant m} \right.\quad = \left. {\nabla _{\,x} ^m \left( \begin{gathered} 3m + 4x \\ 2m + 4x \\ \end{gathered} \right)\;} \right|_{\,x\, = \,0} \quad \ne \quad \left. {\nabla _{\,x} ^m \left( \begin{gathered} 3m + 4x \\ m \\ \end{gathered} \right)\;} \right|_{\,x\, = \,0}$$ but taking the limit as $m\; \to \,\infty$ $$\mathop {\lim }\limits_{m\; \to \,\infty } \sum\limits_{0\, \leqslant \,j\,\left( { \leqslant \,\left\lfloor {m/2} \right\rfloor \, \leqslant \,m} \right)} {\left( { - 1} \right)^j \left( \begin{gathered} m \\ j \\ \end{gathered} \right)\left( \begin{gathered} 3m - 4j \\ 2m - 4j \\ \end{gathered} \right)} \quad \left| {\;0 \leqslant m} \right.\quad = \quad \mathop {\lim }\limits_{m\; \to \,\infty } \left. {\nabla _{\,x} ^m \left( \begin{gathered} 3m + 4x \\ m \\ \end{gathered} \right)\;} \right|_{\,x\, = \,0} = \mathop {\lim }\limits_{m\; \to \,\infty } 4^{\, m}$$ since: $$\left( \begin{gathered} 3m + 4x \\ m \\ \end{gathered} \right) = \frac{1} {{m!}}\left( {3m + 4x} \right)\left( {3m + 4x - 1} \right) \cdots \left( {3m + 4x - m + 1} \right) = \frac{{4^{\,m} }} {{m!}}x^{\,m} + \cdots$$
• Wow, thanks! Just a few comments on small typos in your answer. In the definition of the Backward finite difference there should be $f(x-k)$ right?Also where you show the backward finite difference on a general polynomial, you missed the operator in the middle. Sep 9, 2016 at 16:15
• Furthermore, in the step before you take the limit, the not equal sign should be at the first step and the equal sign on the second, correct? I.e. $\binom{3m+4x}{2m+x}=\binom{3m+4x}{m}$ should still be valid. Sep 9, 2016 at 16:28
• a) fully right, thanks, corrected b)$\binom{3m+4x}{2m+x}=\binom{3m+4x}{m}$ in general for our case (for $3m+4x$ non-negative integer, unless using gamma equivalence, ...), but the deltas are not, since the first is "truncated" due to index excursion. Sep 9, 2016 at 16:55