Short sequence beats long sequence

I have encountered some comparison between two binomial sums. It was amusing how the one with "fewer" summands exceeds (in value) than the other which consists of many more terms. In fact, it even more fascinating to me because in my initial inequality the longer sum had a range $$1\leq k\leq m+n$$, however the experimentation continues to remain positive when I increased the range to $$1\leq k\leq 2m+n$$.

At any rate, I would like to ask:

QUESTION. Assume $$m, n\in \mathbb{Z}_{+}$$. Is this inequality true? $$\sum_{k=1}^n \binom{m + 3n + k - 1}{m + n - 1}\,\,\geq \,\, \sum_{k=1}^{2m+n} \binom{2m + 3n + k - 1}{n-1}.$$

• Note that $$\sum_{k=a}^b \binom{k}{j} = \binom{b+1}{j+1} - \binom{a}{j+1}$$ Mar 28 at 14:33
• True, that is just Pascal's recurrence. Now what? Mar 28 at 14:36
• Now reduce both sums to two terms, and maybe it is easier to compare the two sides of your inequality. Mar 28 at 14:52
• Yes, I tried that ... Mar 28 at 14:55
• You would have saved Rob some time and effort, had you included in your question that you had already tried that approach. What else are you hiding from us? Mar 29 at 2:46

$$\newcommand\bi\binom\newcommand{\tr}{\tilde r}$$In view of Rob Pratt's comment, it is enough to show that $$\begin{equation*} L_m:=A_m-B_m\overset{\text{(?)}}\ge R_m \tag{10}\label{10} \end{equation*}$$ for integers $$m,n\ge1$$, where $$\begin{equation*} A_m:=A_{m,n}:=\bi{m+4n}{m+n},\quad B_m:=B_{m,n}:=\bi{m+3n}{m+n},\quad R_m:=R_{m,n}:=\bi{4m+4n}{n}. \end{equation*}$$

Let $$\begin{equation*} r_m:=\frac{B_m}{A_m},\quad \rho_m:=\frac{R_m}{A_m}, \end{equation*}$$ so that $$L_m=A_m(1-r_m)$$ and $$R_m=A_m\rho_m$$. So, it is enough to show that $$\begin{equation*} r_m+\rho_m\overset{\text{(?)}} \le1. \tag{20}\label{20} \end{equation*}$$

For any integers $$a,b,k$$ such that $$0\le k\le a\le b\ne0$$, $$\begin{equation*} \frac{\bi ak}{\bi bk}=\frac ab\frac{a-1}{b-1}\cdots\frac{a-(k-1)}{b-(k-1)}\le\Big(\frac ab\Big)^k. \end{equation*}$$ So, $$\begin{equation*} r_m\le\Big(\frac{m+3n}{m+4n}\Big)^{m+n}\le\tr_m:=\tr_{m,n}:=\exp-\frac{(m+n)n}{m+4n}. \end{equation*}$$

Next, for $$m,n$$ as before, $$\begin{equation*} \frac{\rho_{m+1}}{\rho_m}-1 =-n\frac pq<0, \end{equation*}$$ where \begin{equation*} \begin{aligned} p&:=22 + 270 m + 920 m^2 + 1184 m^3 + 512 m^4 + 155 n + 1155 m n + 2328 m^2 n \\ &+ 1376 m^3 n + 330 n^2 + 1438 m n^2 + 1328 m^2 n^2 + 265 n^3 + 529 m n^3 + 68 n^4, \end{aligned} \end{equation*} $$\begin{equation*} q:=(1 + 4 m + 3 n) (2 + 4 m + 3 n) (3 + 4 m + 3 n) (4 + 4 m + 3 n) (1 + m + 4 n). \end{equation*}$$ So, $$\rho_m$$ is decreasing in $$m\ge1$$. Also, it easy to see that $$\tr_m$$ is decreasing in $$m\ge1$$.

So, for all $$m\ge1$$ and $$n\ge6$$ \begin{equation*} \begin{aligned} r_m+\rho_m &\le \tr_1+\rho_1 \\ & \begin{aligned} =s(n)&:=\exp\Big(-\frac{(1+n)n}{1+4n}\Big) \\ &+\frac{8 (n+1) (2 n+1) (4 n+3)}{3 (3 n+1) (3 n+2) (3 n+4)} %\le s_6=0.970\ldots <1, \end{aligned} \end{aligned} \tag{30}\label{30} \end{equation*} so that \eqref{20} holds, as desired. The inequality $$s(n)<1$$ for $$n\ge6$$ in \eqref{30} holds because for $$t(n):=(s(n)-1)\exp\frac{(1+n)n}{1+4n}$$ we have $$t(6)=-0.160\ldots<0$$ and $$\begin{equation*} t'(n)=-\frac PQ\,\exp\frac{(1+n)n}{1+4n}<0, \end{equation*}$$ where $$\begin{equation*} P:=176 + 2304 n + 11124 n^2 + 25740 n^3 + 32471 n^4 + 26908 n^5 + 19299 n^6 + 10062 n^7 + 1836 n^8, \end{equation*}$$ $$\begin{equation*} Q:=3 (1 + 3 n)^2 (2 + 3 n)^2 (4 + 3 n)^2 (1 + 4 n)^2, \end{equation*}$$ so that $$t(n)<0$$ for $$n\ge6$$.

The cases of $$n\in\{1,\dots,5\}$$ are substantially easier. For each of these $$5$$ cases, $$r_m+\rho_m$$ is a certain rational expression in $$m$$, and then \eqref{20} fails to hold only for $$(m,n)\in\{(1,1),(2,1),(1,2),(1,3)\}$$ -- but for such $$(m,n)$$ the desired inequality in the OP is checked by a simple direct calculation. $$\quad\Box$$

Too big to comment: LHS$$-$$RHS gives $$f(m,n)=\binom{m+4n}{m+n}-\binom{m+3n}{m+n}+\binom{4m+4n}{n}-\binom{2m+3n}{n}$$.

Now, we can see $$f(0,n)=0$$ and for $$m>>n$$ we get, using Sterling's approximation, $$f(m,n)\approx m^{2n}\left(\frac{m^{n}}{(3n)!}-\frac{1}{(2n)!}\right)-\frac{(2m)^n}{n!}(2^n-1)\approx\frac{m^{3n}}{(3n)!}-\frac{(4m)^n}{n!}>0$$

• Sterling is not only an approximation, but gives rigorous bounds: $n!=(n/e)^n\sqrt{2\pi n} e^{\theta/(12 n}$ for some $\theta\in[0,1]$. So this argument gives a proof for all but a few very small cases. Mar 28 at 19:30
• @Jan-ChristophSchlage-Puchta : I had thought of this approach, but there seemed to be too many factorials to deal with. So, I am surprised to hear that this approach gives a proof. Can you present such a proof? Mar 28 at 20:09
• @Jan-Christoph Schlage-Puchta Thanks for your comment. Mar 29 at 16:41