But does it converge?

Let us first consider if $\displaystyle \lim_{x \to \infty} \sum_{k=0}^n x_k$ converges using the ratio test . Note that $x_n$ is positive for all $n$ .

$\begin{aligned} \lim_{n \to \infty} \frac {x_{n+1}}{x_n} & = \lim_{n \to \infty} \frac {\ln \left(e^{x_n}-x_n \right)}{x_n} < \lim_{n \to \infty} \frac {\ln \left(e^{x_n} \right)}{x_n} = 1 \end{aligned}$

Therefore, $\displaystyle \lim_{x \to \infty} \sum_{k=0}^n x_k$ converges.

Now, we have:

$\begin{aligned} x_{n+1} & = \ln \left(e^{x_n}-x_n \right) \\ e^{x_{n+1}} & = e^{x_n} - x_n \\ \implies x_n & = e^{x_n} - e^{x_{n+1}} \end{aligned}$

Since $\displaystyle \lim_{x \to \infty}\sum_{k=0}^n x_k$ converges, $\implies \displaystyle \lim_{n \to \infty} e^{x_n} = \lim_{n \to \infty} e^{x_{n+1}}$ and $\displaystyle \lim_{x \to \infty} x_n = 0$ .

Then we have:

$\begin{aligned} \sum_{k=0}^\infty x_k & = \lim_{n \to \infty} \sum_{k=0}^n \left(e^{x_k} - e^{x_{k+1}}\right) \\ & = \lim_{n \to \infty} \left(e^{x_0} - e^{x_{n+1}}\right) \\ & = e^1-e^0 = e-1 \end{aligned}$

$\implies A+B = 1+1 = \boxed{2}$