Estimate the error if the sum of sum from n=0 to infinity

Estimate the error if the sum of $\sum_{n=0}^{\infty}\dfrac{x^n}{n!}$ is replaced by the sum of the first $N$ terms.

Explanation of Method

Taylor's theorem with the Lagrange remainder gives an exact expression for the error in terms of an intermediate point $\xi$; since $e^t$ is monotonic, this expression can be bounded in terms of $x$ alone, and for practical purposes the magnitude of the first omitted term already gives an excellent estimate.

Write the remainder $R_N=e^x-\sum_{k=0}^{N-1}x^k/k!$ using Taylor's theorem with Lagrange remainder.
Bound $e^\xi$ using monotonicity of $e^t$.
State the practical first-order estimate and check it numerically.

Step 1 · Lagrange remainder

By Taylor's theorem, for some $\xi$ between $0$ and $x$: $$ R_N = e^x - \sum_{k=0}^{N-1}\frac{x^k}{k!} = \frac{e^{\xi}}{N!}x^N. $$

Step 2 · Bound using monotonicity

For $x>0$: $1

Step 3 · Practical estimate and numerical check

A simple, widely useful first-order estimate is $$ |R_N| \approx \frac{|x|^N}{N!}, $$ since the terms of the series decay factorially fast once $N$ exceeds $|x|$. For $x=1$, $N=10$: $R_{10}\approx1/10!\approx2.76\times10^{-7}$, matching the true error $e-\sum_{k=0}^{9}1/k!\approx2.76\times10^{-7}$ almost exactly.

Summary Table

Property	Value
Exact remainder	$R_N=e^{\xi}x^N/N!$, $\xi$ between $0,x$
Practical estimate	$\|R_N\|\approx\|x\|^N/N!$
Example ($x=1,N=10$)	$\approx2.76\times10^{-7}$

Additional Notes

Because the terms decay factorially rather than geometrically, this series converges dramatically faster than the alternating harmonic series of Problem 279 — for the same number of terms, the error here is astronomically smaller.
The Lagrange remainder form $R_N=e^{\xi}x^N/N!$ is the general template for Taylor remainders of any smooth function; here it simplifies nicely because all derivatives of $e^t$ equal $e^t$ itself.
Once $N$ exceeds $|x|$, each successive term shrinks by a growing factor $|x|/(N+1)$, so the true error is typically noticeably smaller than the crude bound $|x|^N/N!$ — visible in the near-exact numerical match at $x=1,N=10$.

Result

$$ R_N = \frac{e^{\xi}}{N!}x^N \ \ (\xi \text{ between } 0,x), \qquad |R_N| \approx \frac{|x|^N}{N!}. $$