Partial sums of exponential series What is known about $f(k)=\sum_{n=0}^{k-1} \frac{k^n}{n!}$ for large $k$?
Obviously it is is a partial sum of the series for $e^k$ -- but this partial sum doesn't reach close to $e^k$ itself because we're cutting off the series right at the largest terms. In the full series, the $(k+i-1)$th term is always at least as large as the $(k-i)$th term for $1\le i\le k$, so $f(k)< e^k/2$. Can we estimate more precisely how much smaller than $e^k$ the function is?
It would look very nice and pleasing if, say, $f(k)\sim e^{k-1}$ for large $k$, but I have no real evidence for that hypothesis.
(Inspired by this question and my answer thereto).
 A: This appears as problem #96 in Donald J Newman's excellent book: A Problem Seminar. 
The problem statement there is:

Show that
$$ 1 + \frac{n}{1!} + \frac{n^2}{2!} + \dots + \frac{n^n}{n!} \sim
\frac{e^n}{2}$$

Where $a_n \sim b_n$ mean $\lim \frac{a_n}{b_n} = 1$.
Thus we can estimate your sum (I have swapped $n$ and $k$) as
$$ 1 + \frac{n}{1!} + \frac{n^2}{2!} + \dots + \frac{n^{n-1}}{(n-1)!} \sim
\frac{e^n}{2}$$
as by Stirling's formula, $\dfrac{n^n}{n!e^n} \to 0$.
The solution in the book proceeds as follows:
The remainder term for a the Taylor Series of a function $f$ is
$$ R_n(x) = \int_{0}^{x} \frac{(x-t)^n}{n!} f^{n+1}(t) \ \text{d}t$$
which for our purposes, comes out as
$$\int_{0}^{n} \frac{(n-t)^n}{n!} e^t \ \text{d}t$$
Making the substitution $n-t = x$ gives us the integral
$$ \int_{0}^{n} \frac{x^n}{n!} e^{-x} \ \text{d}x$$
In an earlier problem (#94), he shows that
$$\int_{0}^{\infty} \left(1 + \frac{x}{n}\right)^n e^{-x} \ \text{d}x \sim \sqrt{\frac{\pi n}{2}}$$
which using the substitution $n+x = t$ gives
$$ \int_{n}^{\infty} t^n e^{-t} \ \text{d}t \sim \frac{n^n}{e^n} \sqrt{\frac{\pi n}{2}}$$
Using $\int_{0}^{\infty} x^n e^{-x}\  \text{d}x = n!$ and Stirling's formula now gives the result.
To prove that
$$\int_{0}^{\infty} \left(1 + \frac{x}{n}\right)^n e^{-x} \ \text{d}x \sim \sqrt{\frac{\pi n}{2}}$$
He first makes the substitution $x = \sqrt{n} t$ to obtain
$$ \int_{0}^{\infty} \left(1 + \frac{x}{n}\right)^n e^{-x} \ \text{d}x \ = \sqrt{n} \int_{0}^{\infty} \left(1 + \frac{t}{\sqrt{n}}\right)^n e^{-\sqrt{n} t} \ \text{d}t$$
Now $$\left(1 + \frac{t}{\sqrt{n}}\right)^n e^{-\sqrt{n} t}  \le (1+t)e^{-t}$$ and thus by the dominated convergence theorem,
$$ \lim_{n\to \infty} \frac{1}{\sqrt{n}} \int_{0}^{\infty} \left(1 + \frac{x}{n}\right)^n e^{-x} \ \text{d}x $$
$$= \int_{0}^{\infty} \left(\lim_{n \to \infty}\left(1 + \frac{t}{\sqrt{n}}\right)^n e^{-\sqrt{n} t}\right) \ \text{d}t$$
$$ = \int_{0}^{\infty} e^{-t^2/2} \ \text{d}t = \sqrt{\frac{\pi}{2}} $$
A: $e^{-k} f(k) = P(X_k &lt k)$ where $X_k$ is a Poisson random variable with parameter $k$.  Since the sum of $k$ independent Poisson random variables with parameter $1$ is Poisson with parameter $k$, the Central Limit Theorem says the probability distribution of $Z_k = (X_k - k)/\sqrt{k}$ approaches a standard normal distribution.  Thus $P(X_k &lt k) = P(Z_k &lt 0) \to 1/2$ as $k \to \infty$.     
