If $\sum_{k=0}^{n-1}\frac{x^k}{k!}=\sum_{k=n}^{\infty}\frac{x^k}{k!}$ for large $n$, approximate $x$ in terms of $n$. Split the Maclaurin series for $e^x$ into two sub-series - the first $n$ terms, and the remainder - then equate the two sub-series. What is a good approximation for $x$, for large $n$?
Since $e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}$, we have $\sum_{k=0}^{n-1}\frac{x^k}{k!}=\frac{1}{2}e^x$.
Experimenting with desmos, it seems that $x\approx n-\frac{1}{3}$ for even or odd $n$, and $x\approx -0.28n-0.4$ for even $n$. Other than that, I do not know how to approach this question.
(Context: I was trying to answer this question, but I think the question in this post is interesting by itself.)
 A: This is equivalent to the asymptotic inversion with respect to $x$ and for large $n$ the normalised upper incomplete gamma function $Q$: $$
Q(n,x) = \frac{1}{2}.
$$ From this result, we have $$
x \sim n-\frac{1}{3}+\frac{8}{405n}+\frac{184}{25515n^2}+\frac{2248}{3444525n^3}+\ldots,
$$
as $n\to +\infty$. To find the negative root when $n$ is even, we can proceed as follows. For the lower incomplete gamma function $\gamma$, we have
$$
\frac{{\gamma (n + 1,nx)}}{{(nx)^n e^{ - nx} }} = \frac{{n\gamma (n,nx)}}{{(nx)^n e^{ - nx} }} - 1.
$$
Thus, by this result,
$$
\frac{{n\gamma (n,nx)}}{{(nx)^n e^{ - nx} }} = \frac{1}{1-x}+\mathcal{O}\!\left(\frac{1}{n}\right)
$$
for $x<0$ and $n\to +\infty$. For the normalised lower incomplete gamma function this means
$$
P(n,nx)= \frac{{(nx)^n e^{ - nx} }}{{\Gamma (n + 1)}}\frac{1}{1-x}\left( 1+\mathcal{O}\!\left(\frac{1}{n}\right) \right),
$$
for $x<0$ and $n\to +\infty$. Employing Stirling's formula yields
$$
P(n,nx)= \frac{{x^n e^{n(1 - x)} }}{{\sqrt {2\pi n} }}\frac{1}{1-x}\left( 1+\mathcal{O}\!\left(\frac{1}{n}\right)  \right),
$$
for $x<0$ and $n\to +\infty$. Now the problem is equivalent to solving $P(n,nx)=\frac{1}{2}$ for even $n$ and negative $x$. If $n$ is even, the above asymptotics may be written
$$
P(n,nx) = \frac{{e^{n(1 - x + \log \left| x \right|)} }}{{\sqrt {2\pi n} }}\frac{1}{1-x}\left( 1+\mathcal{O}\!\left(\frac{1}{n}\right)\right)
$$
for $x<0$ and $n\to +\infty$. A first approximation to the solution of the original problem is thus
$$
x_0 n=(-0.2784645438\ldots )n,
$$
where $x_0=-0.2784645438\ldots$ is the unique negative root of $1-x+\log|x|=0$. A better approximation follows by solving
$$
(1 - x + \log \left| x \right|) = \frac{1}{{2n}}\log \left( {\frac{{(1-x)^2\pi n}}{2}} \right).
$$
Writing $x = x_0  + \xi$, using linear approximation, and taking $1-x\approx 1-x_0$ on the right-hand side, we find
$$
\frac{{1 - x_0 }}{{x_0 }}\xi  \approx \frac{1}{{2n}}\log \left( {\frac{{(1-x_0)^2\pi n}}{2}} \right) \Longrightarrow \xi  \approx \frac{{x_0 }}{{2(1 - x_0 )n}}\log \left( {\frac{{(1-x_0)^2\pi n}}{2}} \right).$$ Hence, a second approximation to the solution of the original problem is
\begin{align*}
& x_0 n + \frac{{x_0 }}{{2(1 - x_0 )}}\log \left( {\frac{{(1-x_0)^2\pi n}}{2}} \right) \\ & = (-0.2784645438\ldots)n +(-0.1089058528\ldots)\log \left(  (2.5674219652\ldots)n  \right). 
\end{align*}
It is possible to obtain higher approximations using more terms from the asymptotics of $P(n,nx)$, but the procedure becomes tedious.
