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Note that by Cesaro's Theorem, we have as a consequence $$\frac{x_n}n\to 1$$

Consider $$r_n(x)=e^{-x}-\sum_{k=0}^n (-1)^k\frac{x^k}{k!}$$ and $$f_n(x)=(-1)^{n+1}e^{-x}r_n(x)$$

One can argue by $r_n(0)=0$ and $r_{n+1}^\prime=-r_n$ that the $r_n$ alternate signs. Thus $f_n>0$ everywhere but $x=0$ where it is $0$, and $f_n\to 0$ as $x\to\infty$, $n=0,1,2,\ldots$. Thus each $f_n$ has a (unique) local maxima over $[0,\infty)$. But since $$f'_{n+1}=f_n-f_{n+1}$$ the equation $$\Delta f_n(x)=0$$ has a unique solution $x_n$. I want to show that $$\tag{1} \lim_{n\to\infty}\Delta x_n=1$$ $$\tag{2}\lim_{n\to\infty}\sqrt{\alpha n} f_n(x_n)=1$$ for some $\alpha>0$

NOTE Introducing $f_n$ was a related to the linked question. We're effectively looking for roots of $$r_n(x)+r_{n+1}(x)=0$$

which after ignoring the exponential term, reduces to finding the furthest negative solution of $$0 = \sum\limits_{k = 0}^n {\frac{{2{x^k}}}{{k!}}} + \frac{{{x^{n + 1}}}}{{\left( {n + 1} \right)!}}$$

ADD By L'Hopital's rule, noting that $r_n^\prime=-r_{n-1}$ and $$\lim_{x\to\infty}{-r_0(x)}=\lim_{x\to\infty}(1-e^{-x})=1$$ it follows that $$\mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( { - 1} \right)}^{n + 1}}n!{r_n}\left( x \right)}}{{{x^n}}}=1$$

I don't know if this justifies, without better approximations, to look at $$\eqalign{ & {\left( { - 1} \right)^{n + 1}}\frac{{{x^n}}}{{n!}} + {\left( { - 1} \right)^{n + 2}}\frac{{{x^{n + 1}}}}{{\left( {n + 1} \right)!}} = 0 \cr & - 1 + \frac{x}{n} = 0 \Rightarrow x = n \cr} $$

which does confirm the guesses above.

Motivation Weighted uniform convergence of Taylor series of exponential function

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