Truncated exponential series equation

\textbf{Problem}. For a positive integer n let x_n be the unique positive real root of the equation


\sum_{k=0}^n \frac{x^k}{k!}=\frac{e^x}{2}.

Prove that \displaystyle\lim_{n\to\infty}(x_n-n)=\frac{2}{3}.

\textbf{Comment.} The limit \displaystyle\lim_{n\to\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!} is well known to be \displaystyle\frac{1}{2}. It could be derived via nice probabilistic structure related to Poisson distribution. This limit hints that the value of the root x_n should be close to n. 

\textbf{Sketch of a solution of the problem}. Defining 

g(x)=\frac{1}{n!}\int_0^x s^ne^{-s}\ ds-\frac{1}{2} 

the original equation could be rewritten in the form g(x)=0. One should observe that g is monotonically increasing and the unique root is in the interval (n,n+1). Then one uses one iteration of the Newton method initialized at n to obtain an approximation of the solution \tilde x_n. Using the final asymptotic obtained at the end of the this answer and Stirling approximation, one obtains that \displaystyle\lim_{n\to\infty}(\tilde x_n-n)=\frac{2}{3}. It remains to use the estimate of the error in the Newton method (one needs the first and second derivatives of g here, which are easy to write), to see that \lim_{n\to\infty}(x_n-\tilde x_n)=0.

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