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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