Le Cam's theorem and total variation distance Le Cam's theorem gives the total variation distance between the sum of independent Bernoilli variables and a Poisson random variable with the same mean.  In particular it tells you that the sum is approximately Poisson in a specific sense.
Define
$$S_n = X_1+\dots+X_n \text{ and } \lambda_n = p_1+\dots+p_n$$
where $P(X_i = 1) = p_i$.
The theorem states that
$$\sum_{k=0}^{\infty}\left| P(S_n=k)-\frac{\lambda_n^k e^{-\lambda_n}}{k!}\right| < 2\sum_{i=1}^n p_i^2.$$
I am having problems understanding what this tells you about the relationship between their cdfs. That is between $P(S_n < x)$ and $P(\operatorname{Poiss}(\lambda_n)) < x)$.   In particular, can you given $n$ and $x$ give a bound on the difference or can you say that as $n$ grows the difference tends to zero?

Let $Y_n$ be any Poisson random variable with parameter $\lambda_n$. Then, for every $x$,
$$
\left|P(S_n < x)-P(Y_n < x)\right|=\left|\sum_{k\lt x} P(S_n=k)-P(Y_n=k)\right|\leqslant\sum_{k=0}^{\infty}\left| P(S_n=k)-P(Y_n=k)\right|.
$$
Hence,
$$
\left|P(S_n < x)-P(Y_n < x)\right| < 2\sum_{i=1}^n p_i^2.$$
