Upper bound for $\mathbb{E}[1/(1+M)]$, where $M$ follows Binomial distribution $B(n,p)$

Question

Since $\frac{1}{1+M}$ is convex, it's easy to get $\mathbb{E}[\frac{1}{1+M}] \ge \frac{1}{\mathbb{E}[M]+1} = \frac{1}{np+1}$. However, what I need is one simply upper bound in terms of $n^{-1}$. Any good idea? Upper bound in the case that $n$ is big enough is also ok.

Answer 1

According to this answer, $\mathbb{E}(\frac{1}{1+M}) =\frac{1}{(n+1)p}\cdot(1-(1-p)^{n+1})$, the upper bound can be $\frac{1}{(n+1)p}$.

Answer 2

Heuristics

You may play with different values of $n$ in my online graph. (I have changed $p$ to $x$ for technical reasons, replaced $n$ with $10^n$ (injective function on $n$), and $\log$ preserves boundedness on $p\in(0,1)$). The $y$-intercept of the strictly decreasing convex curve is observed to be the value of $n$ that you've set.

Motivation (edited in response to comments)

Based on another answer, $$\bbox[2px, border: 1px solid black]{\mathbb{E}\left[\frac{1}{1+M}\right] =\frac{1}{n+1}\cdot\frac{1-(1-p)^{n+1}}{p}} \color{gray}{\left(\le\frac{1}{(n+1)p}\right)}.$$
If $p$ becomes $\text{s}\small\text{ma}\tiny\text{ll}$, then $\left((n+1)\tiny{p}\right)^{\Large{-1}} \large\text{blows } \Large\text{up}$. Since OP asks for an upper bound in terms to $n$, I'll try to get rid of $p$ by taking supremum on $p\in(0,1)$. By doing so, we will see eventually that the upper bound is $1$. From the first comment, you will see that this is immediate from monotonicity of $\Bbb E$. However, by doing so, you will know that the best upper bound for $\Bbb {E}[1/(1+M)]$ independent of $p$ is $1$.

Formal derivation

From the graph, we see that the key step is to establish monotone convergence $(1-(1-p)^{n+1})/p \uparrow n$ as $p \downarrow 0$ for each fixed $n$. (Monotonicity allows us to put an inequality sign without considering $\epsilon$-$\delta$.)

To get rid of $p$, for each fixed $n$, define \begin{align} f_n(p)&=\frac{1-(1-p)^{n+1}}{p} \quad \forall\,p\in(0,1). \\ f_n'(p)&=\frac{(n+1)p(1-p)^n-(1-(1-p)^{n+1})}{p^2} \\ &= \frac{[(n+1)p+(1-p)](1-p)^n-1}{p^2} \\ &= \frac{(np+1)(1-p)^n-1}{p^2} < 0 \end{align}

The last inequality is due to the fact that the range of $g_{jn} (x) =(1+jx)(1-x)^n$ is $[0,1]$. In the answer of that question, $g_{jn} '(x) < 0$, so $g_{jn} (x)$ is injective. In this case, take $j=n$ and $x=p$. Since we suppose that $p \in (0,1)$, we can take away the the endpoints $0$ and $1$ from the range and conclude that $0<g_{nn}(p)=f_n(p)<1$ for all $p \in (0,1)$.

To evaluate the limit of $f_n(p)$ as $p\downarrow0$, we expand the first few terms of the denominator. The terms with degree $\ge2$ with respect to $p$ vanish as $p\downarrow0$, leaving $n+1$ alone.

$$\bbox[2px, border: 1px solid black]{f_n(p) \le \frac{1-(1-p)^{n+1}}{p} \le n+1}$$

Combining this with another answer, we have $$\mathbb{E}\left[\frac{1}{1+M}\right] \le 1.$$ No better upper bound independent of $p$ exists.