Does either/both of the properties $E[M]=mp$ and $M\stackrel{a.s.}{\to}mp$ imply that $E\left[\frac{1}{M}\right]=\frac{1}{mp}$?

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Let $X_1,\dots,X_m$ be a set of $m$ i.i.d observations. Define $$ M = \sum_{i=1}^m 1_{X_i \in [a,b]}. $$ where $1_{X_i \in [a,b]}$ is the indicator function for the observation $X_i$ falling in the interval $[a,b]$. Define $p$ to be the probability of an observation falling the interval $[a,b]$. $$ p = P(X_i \in [a,b]) = E[1_{X_i \in [a,b]}]. $$

Note that $E[M] = mp$, and $M \stackrel{a.s.}{\to}mp$.

Since we have these properties for $M$, I want to know can we say $$ \begin{align} E\bigg[\frac{1}{M}\bigg] = \frac{1}{mp}, \quad \quad \text{and} \quad \quad E\bigg[\frac{1}{M^2}\bigg] = \frac{1}{(mp)^2}. \end{align} $$

If this is not valid, does it hold that $$ \begin{align} E\bigg[\frac{1}{M}\bigg] \to \frac{1}{mp}, \quad \quad \text{and} \quad \quad E\bigg[\frac{1}{M^2}\bigg] \to \frac{1}{(mp)^2}, \end{align} $$ in some appropriate mode of convergence as $m \to \infty$?