Somemathhere

$\textbf{Problem}.$ 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 a