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?
> $$\sum\_{k=1}^n \binom{m + 3n + k - 1}{m + n - 1}\,\,\geq \,\,
> \sum\_{k=1}^{2m+n} \binom{2m + 3n + k - 1}{n-1}.$$
> 
> 
> 



Short sequence beats long sequence
$\newcommand\bi\binom\newcommand{\tr}{\tilde r}$In view of [Rob Pratt's comment](https://mathoverflow.net/questions/443677/short-sequence-beats-long-sequence#comment1145361_443677), 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. $\quad\Box$


