Limit of a sum. While fixing my answer to this question I noticed that (actually the question is equivalent to this modulo some algebra)
$$\frac{1}{2}=\lim_{x\to\infty}\sum_{i=0}^\infty \frac{x^{i-1}}{i!\sum_{j=0}^i\frac{x^j}{j!}}$$
The series converges pretty much like an exponential series, so is easy to numerically evaluate, but I cannot seem to beat it into submission.
Further experimentation leads me to conjecture that
$$\sum_{i=0}^\infty \frac{x^{i}}{i!\sum_{j=0}^i\frac{x^j}{j!}} = \frac{x+\ln x}{2}+O(1)$$
 A: Here is a very rough heuristics: Let $S(x)$ denote the sum
$$ S(x) = \sum_{n=0}^{\infty} \frac{x^n/n!}{\sum_{j=0}^{n} x^j / j!} \cdot \frac{1}{x}. $$
Using the falling factorial $(n)_k := n(n-1)\cdots(n-k+1) = n!/(n-k)!$ we can rewrite $S(x)$ as
$$ S(x) = \sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} \frac{(n)_k}{x^k} \right)^{-1} \frac{1}{x}. $$
Regarding $S(x)$ as a kind of Riemann sum, $S(x)$ approximates the integral
$$ \int_{0}^{\infty} \frac{1}{1+t+t^2+\cdots} \, dt = \int_{0}^{1} (1-t) \, dt = \frac{1}{2}. $$
This explains why we expect the limit as $1/2$. Turning this argument into a more rigorous form is not that hard. Indeed, we prove this by showing that both limsup and liminf equals $1/2$.
Estimating Liminf. Let $a \in (0, 1)$. Then for $n < x$ we have
$$ \sum_{k=0}^{n} \frac{(n)_k}{x^k}
\leq \sum_{k=0}^{n} \left( \frac{n}{x} \right)^k
\leq \sum_{k=0}^{\infty} \left( \frac{n}{x} \right)^k
= \left(1 - \frac{n}{x}\right)^{-1} $$
and hence
$$ S(x)
\geq \sum_{n \leq a x} \left( \sum_{k=0}^{n} \frac{(n)_k}{x^k} \right)^{-1}
\geq \sum_{n \leq a x} \left( 1 - \frac{n}{x} \right) \frac{1}{x}.
$$
This lower bound is a Riemann sum, thus taking limsup as $x \to \infty$ yields
$$ \liminf_{x\to\infty} S(x) \geq \int_{0}^{a} (1 - t) \, dt. $$
Since the LHS does not depend on $a$, taking $a \uparrow 1$ gives
$$ \liminf_{x\to\infty} S(x) \geq \frac{1}{2}. $$
Estimating Limsup. Let $b > 1$ and $X \sim \operatorname{Poisson}(x)$. Then by the law of large numbers, we easily see that $\Bbb{P}(X \leq bx) \to 1$ as $x \to \infty$. So we have
\begin{align*}
\sum_{n > bx} \frac{x^n/n!}{\sum_{j=0}^{n} x^j / j!} \cdot \frac{1}{x}
&= \frac{1}{x} \sum_{n > bx} \frac{\Bbb{P}(X = n)}{\Bbb{P}(X \leq n)} \\
&\leq \frac{1}{x} \frac{\Bbb{P}(X > bx)}{\Bbb{P}(X \leq bx)} \\
&\longrightarrow 0 \quad \text{as } x \to \infty.
\end{align*}
This observation allows us to truncate the sum without affecting the limsup. Now let $N$ be any positive integer. Then focusing on the summands $n$ in the region $ N \leq n \leq bx$, we get
\begin{align*}
S(x)
&= \sum_{N \leq n \leq bx} \left( \sum_{k=0}^{n} \frac{(n)_k}{x^k} \right)^{-1} \frac{1}{x} + o(1) \\
&\leq \sum_{N \leq n \leq bx} \left( \sum_{k=0}^{N} \frac{(n)_k}{x^k} \right)^{-1} \frac{1}{x} + o(1).
\end{align*}
This upper bound is again a Riemann sum. Taking limsup as $x \to \infty$, we get
$$ \limsup_{x\to\infty} S(x) \leq \int_{0}^{b} \frac{dt}{1 + t + \cdots + t^N}. $$
Again, since $b$ and $N$ are independent of the LHS, taking $b \downarrow 1$ and $N \to \infty$ gives
$$ \limsup_{x\to\infty} S(x) \leq \int_{0}^{1} \frac{dt}{1 + t + \cdots} = \frac{1}{2}.$$
Therefore the conclusion follows.
