I think this problem will help me learn some convergence techniques.
Given $(\xi_j)_{j=1}^{\infty} \sim \xi$. Supposse that in probability $$\frac{1}{n}\sum_{j=1}^n \exp ( i s \xi_j ) \to E[\exp ( i s \xi )]\quad (n \to \infty)$$
I want to show that
$$ \Pi_{j=1}^n \left[ 1 + \frac{\lambda}{n} \Big( \exp (i s \xi_j) - 1 \Big) \right] \to \exp( \lambda ( E[ \exp (i s \xi) ] - 1)) \quad (n \to \infty)$$
My attempt is: denote $X_{j}^n = \left[ 1 + \frac{\lambda}{n} \Big( \exp (i s \xi_j) - 1 \Big) \right]$, $X_n = \Pi_{j=1}^n X_j^n $ and try to show that $$\log X_n \to \lambda ( E[ \exp (i s \xi) ] - 1)$$
So take the log
$$\hat{X}_n := \log X_n = \sum_{j=1}^n \log X_j^n = \sum_{j=1}^n \log \left[ 1 + \frac{\lambda}{n} \Big( \exp (i s \xi_j) - 1 \Big) \right] $$
But I don't know how to conclude.
Help
Note that $x-\frac{x^2}{2}\leq\log(1+x)\leq x$, thus $$(X_j^n-1)-\frac{(X_j^n-1)^2}{2}\leq\log(1+(X_j^n-1))\leq (X_j^n-1).$$
For the RHS, by SLLN we have $$\sum_{j=1}^n(X_j^n-1)=\frac 1n\sum_{j=1}^n\lambda(e^{is\xi_j}-1)\to\lambda(\mathbb{E}e^{is\xi}-1).$$
For the LHS we simply note that the quadratic term $$|(X_j^n-1)^2|=\frac{\lambda^2}{n^2}|e^{is\xi_j}-1|^2\leq\frac{\lambda^2}{n^2}\cdot4$$
sums up to $\frac{C}{n}\to 0$ for some $C$. This proves the result with a.s. convergence.