Convergence of a Branching process Consider the Branching process: $\{ \xi_i^n , n \ge 1, i \ge 1\}$ are i.i.d. taking values $0, 1, \ldots$ and $Z_0 := 1, \; Z_{n+1} := \sum\limits_{i=1}^{Z_n} \xi_i^{n+1}$. Assume $\mu := \mathbb{E}[\xi_i^b]>1$. Assume $\sigma^2 := \operatorname{Var}(\xi_i^n) < \infty$. Denote $X_n := \frac{Z_n}{\mu^n}$. I need to show that $\lim\limits_{n \longrightarrow \infty} \mathbb{E}[|X_n - X_\infty|^2] = 0$ and that $P(X_\infty > 0) > 0$.
I can show that $X_n$ defined as such, is a martingale and that $X_n \longrightarrow X_\infty$ a.s. for some random variable $X_{\infty}$. Furthermore, I derived $\mathbb{E}[|X_n - X_{n-1}|^2] = \frac{\sigma^2}{\mu^{n+1}}$. Since $X_n$ is a martingale, I can interpret $\mathbb{E}[|X_n - X_{n-1}|^2]$ as the conditional variance $\operatorname{Var}(X_n \mid X_{n-1})$. I am trying to use this latter result to prove the convergence. 
I appreciate any insights on this.

Suppose the conditional expectation of $X_n$ given $X_{n-1}$ is $X_{n-1}$, as in a martingale, and the conditional variance is $2$.  Then the law of total variance tells us that
$$
\operatorname{var}(|X_n - X_{n-1}|^2) = \operatorname{var}(\operatorname{E}(|X_n-X_{n-1}|^2 \mid X_{n-1}) + \operatorname{E}(\operatorname{var}(|X_n-X_{n-1}|^2 \mid X_{n-1}). \tag 1
$$
In cases where the conditional variance of $X_n$ given $X_{n-1}$ does not depend on $X_{n-1}$, the first term on the right side in $(1)$ is the conditional variance of $X_n$ given $X_{n-1}$, and typically the second term is positive, not $0$.  Therefore the expression on the left side is not the conditional variance of $X_n$ given $X_{n-1}$, but is a larger quantity.
