What is $\mathbb{E}[(1 - (1 - p)^{X})(n-X-1)]$ as $n \rightarrow \infty$? Let $X \sim \text{Bin}(n-1,p)$ and $p = \frac{\lambda}{n}$. What is $\mathbb{E}[(1 - (1 - p)^{X})(n-X-1)]$ as $n \rightarrow \infty$ ?
Here is my attempt:
\begin{align}
\mathbb{E}[(1 - (1 - p)^{X})(n-X-1)] &= \mathbb{E}[(n-1) - X - (n-1)(1 - p)^{X} + X(1 - p)^{X}]\\
&= (n-1) -\mathbb{E}[X] - (n-1)\mathbb{E}[(1 - p)^{X}] + \mathbb{E}[X(1 - p)^{X}].
\end{align}
\begin{align}
\mathbb{E}[X] &= (n-1)p\\
\mathbb{E}[(1 - p)^{X}] &= \mathbb{E}[e^{\ln{(1 - p)^{X}}}] = \mathbb{E}[e^{X\ln{(1-p)}}] = (1 - p + pe^{\ln{(1-p)}})^{n-1} \hspace{15mm} (\text{by momemnt generating function})\\
&= ((1+p)(1-p))^{n-1}\\
\mathbb{E}[X(1 - p)^{X}] &= \sum_{k=0}^{n-1}k(1-p)^{k}{n-1 \choose k}p^{k}(1-p)^{(n-1)-k} = \sum_{k=1}^{n-1}k(1-p)^{k}{n-1 \choose k}p^{k}(1-p)^{(n-1)-k} \\
&=(n-1)p\sum_{\ell=0}^{n-2}(1-p)^{\ell+1}{n-2 \choose \ell}p^{k-1}(1-p)^{(n-2)-\ell} = (n-1)p(1-p)\mathbb{E}[(1 - p)^{Y}]\\
&= (n-1)p(1-p)((1+p)(1-p))^{n-2}, \hspace{5mm} \text{since } Y \sim \text{Bin}(n-2,p).
\end{align}
Now putting it together, I believe, we obtain:
\begin{align}
\mathbb{E}[(1 - (1 - p)^{X})(n-X-1)] = (n-1)\left[(1 - p) - ((1+p)(1-p))^{n-1} + p(1-p)((1+p)(1-p))^{n-2}\right].
\end{align}
Focusing on the term in the square brackets. As $n \rightarrow \infty$,
\begin{align}
(1-p) &= 1-\tfrac{\lambda}{n} \rightarrow 1 \\
((1+p)(1-p))^{n-1} &= (1+\tfrac{\lambda}{n})^{n-1}(1-\tfrac{\lambda}{n})^{n-1} \rightarrow e^{\lambda}e^{-\lambda} = 1 \\
((1+p)(1-p))^{n-2} &= (1+\tfrac{\lambda}{n})^{n-2}(1-\tfrac{\lambda}{n})^{n-2} \rightarrow 1 \\
p(1 -p) &= \tfrac{\lambda}{n}(1 - \tfrac{\lambda}{n}) \rightarrow 0 \\
\end{align}
So the term in the square brackets converges toward zero. Moreover, $(n-1) \rightarrow \infty$. This leaves me with the question where the whole term converges toward?
Important: according to my lecture notes the expectation should converge toward $\lambda^2$. The exercise is related to the relationship between a offspring process and the Erdos-Renyi random graph.
Also, this is my first question asked on stack exchanged. I really hope I followed the guidelines of the platform correctly. If not, my apologies. Please let me know how to improve my question :D
 A: Note that
\begin{align}
&(n-1)[(1-p)-((1+p)(1-p))^{n-1}+p(1-p)((1+p)(1-p))^{n-2}]\\
=&(1-p)(n-1)[ 1-((1+p)(1-p))^{n-2}(p-(1+p))]\\
=&(1-p)(n-1)[1-(1-p^2)^{n-2}]
\end{align}
Now, $(1-p)$ just converges to $1$, so we'll ignore that. Now, by the binomial formula
\begin{align}
(n-1)[1-(1-p^2)^{n-2}] &=-(n-1)\sum_{k= 1}^{n-2} {n-2 \choose k} (-p^2)^k\\
&= (n-1)(n-2)p^2\\
&-(n-1)\sum_{k= 2}^{n-2} {n-2 \choose k} (-p^2)^k
\end{align}
The first term converges to $\lambda^2$, so we have to prove that the last term converges to $0$. For this, note that ${ \alpha \choose \beta+1}/{\alpha\choose \beta}=\frac{\alpha-\beta}{\beta+1}$. And hence, for $n\geq 4$,
\begin{align}
\left|(n-1) \sum_{k=2}^{n-2}{n-2 \choose k} (-p^2)^k\right|&\leq (n-1)p^2 \sum_{j=1}^{n-3} { n-2\choose j} \frac{n-2-j}{j+1} p^{2j}\\
&\leq (n-1)p^2\sum_{j=1}^{n-3} {n-2\choose j} \frac{1}{j+1}\left(\frac{\lambda^2}{n}\right)^j
\end{align}
Now, $(n-1)p^2$ converges to $0$ and, applying the binomial formula and elementary properties of the exponential function (more precisely, convexity and the fundamental equation),
$$
\sum_{j=1}^{n-3} {n-2\choose j} \frac{1}{j+1}\left(\frac{\lambda^2}{n}\right)^j\leq \left(1+\frac{\lambda^2}{n}\right)^{n-2}\leq \exp(\lambda^2)\left(1+\frac{\lambda^2}{n}\right)^{-2}\leq \exp(\lambda^2),
$$
All in all, we've established the desired.
