Some problems appear easy on the face of it, but perhaps they are not. Here is an instance of a certain calculation which is slightly reformulated from its original encounter in a current work. I have no proof.
QUESTION. Assume we are working in a domain where the denominators do not vanish. Is this true? $$\sum_{k=0}^{2t}\frac{(-1)^k\, \binom{q+1}k}{\binom{n+k-q-1}{n-m-q}}\geq0.$$
Let $N$ be a positive integer and $c_1,\ldots,c_N$ non-negative real numbers. Denote $f(x)=((x+c_1)\ldots (x+c_N))^{-1}$.
Lemma 1. For all integer $d\geqslant 0$ and all $x>0$ we have $(-1)^df^{(d)}(x)>0$.
Proof. Induction in $d$. Base $d=0$ is trivial. For doing an induction step it suffices to note that $-f'$ is a sum of $N$ functions of the same type.
Lemma 2. For a non-negative integer $d$, real $a$, and a continuous real function $f(x)$ defined on the segment $[a,a+d]$ and having $d$ derivatives on the interval $(a,a+d)$ ($d$-th derivative may be discontinuous) there exists $\theta\in (0,d)$ such that $$f^{(d)}(a+\theta)=\sum_{k=0}^d (-1)^{d-k}{d\choose k}f(a+k)=:c$$
Proof. Denote $h(x)=\sum_{k=0}^d f(a+k){x\choose k}{d-x\choose d-k}$.Then $h(x)=\frac{c}{d!}x^d+\ldots$ is a polynomial of degree at most $d$ and $h(k)=f(a+k)$ for all $k=0,1,\ldots,d$. Therefore, by Rolle's theorem, the function $g(x):=f(a+x)-h(x)$, which has roots at points $0,1,\ldots,d$, satisfies $g^{(d)}(\theta)=0$ for certain $\theta\in (0,d)$. This $\theta$ is what we need.
Now call a function $f(k)$ good if it has a type $f(k)=\frac{C}{(k+r)(k+r+1)\ldots (k+s)}$ for certain $C>0$ and positive integers $r\leqslant s$. These two lemmata already yield that when $q+1\leqslant 2t$, then $\sum_{k=0}^{2 t} (-1)^k{q+1\choose k}f(k) \geqslant 0$ that is your claim (is it?)
For larger values of $q$, we induct on $q$ with use of Pascal's recurrence ${q+1\choose k}={q\choose k}+{q\choose k-1}$ which allows to write \begin{align*} \sum_{k=0}^{2t} (-1)^k {q+1\choose k}f(k) &=\sum_{k=0}^{2t} (-1)^k {q\choose k}f(k)-\sum_{k=0}^{2t-1} (-1)^k{q\choose k}f(k+1) \\ &={q\choose 2t}f(2t+1)+\sum_{k=0}^{2t}(-1)^k{q\choose k}(f(k)-f(k+1)). \end{align*} Since the function $k\to f(k)-f(k+1)$ is good, we reduced $q+1$ to $q$.
I think one can simplify Fedor's proof using the point of view of Bernstein's Theorem for completely monotone functions, together with Bonferroni's Inequalities from probability.
The given sum is $$ S=(n-m-q)!\ \sum_{k=0}^{2t}(-1)^k\binom{q+1}{k}\ \frac{1}{(k+m)(k+m+1)\cdots(k+n-q-1)} $$
$$ =(n-m-q)!\int_0^{\infty}\left(\sum_{k=0}^{2t} (-1)^k\binom{q+1}{k} e^{-kx}\right)\ d\mu(x) $$ where $\mu$ is the convolution of the measures on $[0,\infty)$ given by $e^{-mx}dx$, $e^{-(m+1)x}dx,\ldots,\ e^{-(n-q-1)x}dx$. All that matters is that $\mu$ is a positive measure. Then take $y=e^{-x}$ in the following lemma.
Lemma:
For all $y\in [0,1]$, $$ \sum_{k=0}^{2t}\binom{q+1}{k} (-y)^k\ge (1-y)^{q+1}. $$
For the proof see the nice write-up by user Mike Earnest
https://math.stackexchange.com/a/4490819/244562
in relation to the Bonferroni Inequalities. I think one can also use Abel (discrete) summation by parts and reduce to the partial alternate sums of binomials which, if I remember correctly, give a sign times a binomial.
