On a sum of binomial coefficients and inequalities

Question

Let $n\ge d\ge 3$ be positive integers.

Is there a closed form formula for $\sum_{i=0}^d \binom {n-d+i-1}{i}$ ? For what conditions on $n$ and $d$ can we say

$\sum_{i=0}^d \binom {n-d+i-1}{i} \le (n-d+1)(d-1)!$ ?

Answer 1

$$ \begin{align} \sum_{i=0}^d\binom{n-d+i-1}{i} &=\sum_{i=0}^d\binom{n-d+i-1}{i}\binom{d-i}{d-i}\tag1\\ &=(-1)^d\sum_{i=0}^d\binom{d-n}{i}\binom{-1}{d-i}\tag2\\ &=(-1)^d\binom{d-n-1}{d}\tag3\\ &=\binom{n}{d}\tag4 \end{align} $$ Explanation:
$(1)$: multiply by $[d\ge i]$
$(2)$: apply negative binomial coefficients
$(3)$: Vandermonde's Identity
$(4)$: apply negative binomial coefficients

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$\binom{n}{d}\le(n-d+1)(d-1)!$ is equivalent to $$ \binom{n}{d-1}\le d!\tag5 $$ For a rough approximation of $n$ in $(5)$, note that $\binom{n}{d-1}\le\frac{n^{d-1}}{(d-1)!}$, then we get that $(5)$ is true when $$ n\le\left(d!(d-1)!\right)^{\frac1{d-1}}\sim\frac{d^2}{e^2}\tag6 $$ where the asymptotic approximation comes from Stirling's Formula.