Let $X$ be binomial distributed with parameters $(n, p)$.
We can easily prove $\mathbb{E} \Big[\frac{n+1}{n+1-X}\Big]$ is an increasing function of $n$ by breaking the combinatorial.
Now what really puzzles me is how to prove $\mathbb{E}\Big[ \text{min}\Big(c, \frac{n+1}{n+1-X}\Big)\Big]$ is also an increasing function of $n$ where c is a constant. i.e.
$$ f(n) = \sum_{m=0}^{n} \min \Big(\frac{n+1}{n+1-m}, c\Big) \binom{n}{m} p^n (1-p)^{n-m} $$
is an increasing function of $n$.
I am thinking of using the Stochastic ordering property, that is if $A\preceq B$ if and only if for all non-decreasing functions $u(.)$, $\mathbb{E}[u(A)] \leq \mathbb{E}[u(B)]$.
Let $X \sim \text{Binomial} (n, p)$ and $Y \sim \text{Binomial} (n+1, p)$. We know $Y$ stochastically dominates $X$ as the former has one more trial. But comparing $(\frac{n+1}{n+1-X})$ with $(\frac{n+2}{n+2-Y})$ does not look very obvious.
Any thought?