Putnam Series

We'll begin by assuming( the proof is left to the reader)that the series is convergent.
Now, $\large x_{n+1} = \ln(e^{x_{n}}-x_{n})$ Or, Equivalently: $\large e^{x_{n+1}} = e^{x_{n}}-x_{n}$ $\large e^{x_{n+1}} = e^{x_{n-1}}-x_{n-1}-x_{n}$ $\large e^{x_{n+1}} = e^{x_{n-2}}-x_{n-2}-x_{n-1}-x_{n}$ Continuing like this, we see that: $\large e^{x_{n+1}} = e^{x_{0}} - \left( x_{0}+x_{1}+x_2+x_3+\cdots+x_n \right)$ Or, $\large e^{x_{n+1}} = e - \sum_{i=0}^{n}x_i$ Or, $\large \sum_{i=0}^{n}x_i = e- e^{x_{n+1}}$ As, the series is convergent, it implies $\large \lim_{n \rightarrow \infty}x_n = 0$ Hence, $\large \lim_{n\rightarrow \infty} \left( \sum_{i=0}^{n}x_i \right) = e-e^0$ Or, $\large \sum_{i=0}^{\infty}x_i = e-1 \approx \boxed{\boxed{ 1.718281828459045}}$

Proof of convergence of the series:

From the equation : $e^{x_{n+1}}=e^{x_n} - x_n$

$e^{x_n} - e^{x_{n+1}}=x_n$

By induction it can be shown that $x_n > 0$ , So , $e^{x_n} >e^{x_{n+1}}$ That implies $x_n > x_{n+1}$

Using exponential series $e^{x_n}=1 +\frac{x_n}{1} +\frac{x_{n}^2}{2!} + \cdots$

$e^{x_n} >1+x_n \\ \implies e^{x_n} -x_n >1 \\ \implies \ln(e^{x_n} -x_n) >0 \\ \implies x_{n+1} >0$

So, the series is decreasing and it is bounded below So it is convergent sequence.and its limiting value is $0$ and then @Upanshu Gupta 's solution.