# 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).

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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,

$$\lim_{n\to \infty} \frac{1}{\sqrt{n}} \int_{0}^{\infty} \left(1 + \frac{x}{n}\right)^n e^{-x} \ \text{d}x$$

$$= \int_{0}^{\infty} \left(\lim_{n \to \infty}\left(1 + \frac{t}{\sqrt{n}}\right)^n e^{-\sqrt{n} t}\right) \ \text{d}t$$

$$= \int_{0}^{\infty} e^{-t^2/2} \ \text{d}t = \sqrt{\frac{\pi}{2}}$$

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$e^{-k} f(k) = P(X_k < k)$ 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 $P(X_k < k) = P(Z_k < 0) \to 1/2$ as $k \to \infty$.