A calculus problem by Akeel Howell

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

Therefore, we have

$\begin{aligned} A & = 1 + \sum_{N=0}^\infty \ln \left(e-\sum_{n=0}^N x_n \right) \\ & = 1 + \sum_{N=0}^\infty \ln \left(e-\sum_{n=0}^N \left(e^{x_n} - e^{x_{n+1}}\right)\right) \\ & = 1 + \sum_{N=0}^\infty \ln \left(e-e^{x_0} + e^{x_{N+1}}\right) \\ & = 1 + \sum_{N=0}^\infty \ln \left(e-e^1 + e^{x_{N+1}}\right) \\ & = 1 + \sum_{N=0}^\infty \ln \left(e^{x_{N+1}}\right) \\ & = 1 + \sum_{N=0}^\infty x_{\color{#3D99F6}N+1} \\ & = \lim_{M \to \infty} \sum_{N=0}^M x_{\color{#D61F06}N} \\ & = \lim_{M \to \infty} \left( e^{x_0} - e^{x_{M+1}} \right) & \small \color{#3D99F6} \text{See note: }\lim_{k \to \infty} x_{k} = 0 \\ & = e - 1 \approx 1.718 \end{aligned}$

$\implies \lfloor 1000A\rfloor = \boxed{1718}$

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Note: From $x_{n+1} = \ln \left(e^{x_n} - x_n \right) < \ln \left(e^{x_n} \right) = x_n$ . This implies that $x_n$ is decreasing from $x_0 = 1$ and since $\ln \left(e^{x_n} - x_n \right) \ge 0$ , $\displaystyle \implies \lim_{n \to \infty} x_n = 0$ .

We have that $x_0 = 1$ and $x_{n+1} = \ln{(e^{x_n} - x_n)}$ . Thus, the recursive definition of the sequence gives

$x_0 = 1 \ \implies \ x_1 = \ln{(e-1)} \ \implies \ x_2 = \ln{(e-1-\ln{(e-1)})} \ \implies \ x_3 = \ln{(e-1-\ln{(e-1)} - \ln{(e-1-\ln{(e-1)})})} \ \cdots$ .

Hence, we see that $x_{n+1} = \ln{(e - x_0 - x_1 - x_2 - \cdots - x_n)}$ .

On expansion of $A$ , we see that $\displaystyle \sum_{N = 0}^{\infty}{\ln{\left( e - \sum_{n = 0}^N{x_n} \right)}} = \ln{(e-1)} + \ln{(e - (1+\ln{(e-1)}))} + \ln{(e - (1 + \ln{(e-1)} + \ln{(e - 1 - \ln{(e - 1)}})))} + \cdots$ which is the sum of all the terms of the sequence, with the exception of the first.

Since $x_{n+1} = \ln{(e^{x_n} - x_n)}$ , we have that $\ \ln{(e^{x_n} - x_n)} = \ln{(e - x_0 - x_1 - x_2 - \cdots - x_n)}$ .

So $\displaystyle x_0 + x_1 + x_2 + \cdots = \sum_n x_n = 1+\sum_n {\ln{(e^{x_n} - x_n)}} = e - \lim_{n \to \infty}{e^{x_{n+1}}} = e - 1$ .

$\therefore 1 + \displaystyle \sum_{N = 0}^{\infty}{\ln{\left( e - \sum_{n = 0}^N{x_n} \right)}} = 1 + e - 2 = e - 1$ .

Thus, $\lfloor {1000A} \rfloor = \ \boxed{1718}$ .