Calculating exponential function with Maclaurin - when stop? I want to calculate $e^x$ using $e^x =\sum_{n=0}^{\infty} \frac{x^n}{n!}$. How much terms i have to calculate if i want to stop calculating when $\frac{x^n}{n!} < 0.01$?
We are searching for the biggest $n$ such that $\frac{x^n}{n!} < 0.01$ and now I'm stucked... 
 A: I think the question to be more fruitful should be in this  way: from any integer $n$ the error of the approximation 
$$e^x \approx\sum_{k=0}^
n \frac{x^k}{k!}$$
is less than $0.01$?
thus we look for $n$ such that the remainder
$$R_n=\sum_{k=n+1}^\infty \frac{x^k}{k!}<0.01$$
Let's take an example with $x=3$, so we have for $k$ sufficiently large:
$$R_n=\sum_{k=n+1}^\infty \frac{3^k}{k!}\leq\sum_{k=n+1}^\infty\frac{3^k}{4^k}=\left(\frac{3}{4}\right)^{n+1}\frac{1}{1-3/4}=4\left(\frac{3}{4}\right)^{n+1}<0.01\iff n+1>\frac{\log 400}{\log4/3}\approx20.8$$
hence we can conclude that with $n=20$,
$$e^3\approx\sum_{k=0}^{20}\frac{3^k}{k!}$$
is an approximation of $e^3$ with an error less than $0.01$.
Numerically we find with Maple $e^3\approx 20.086$ and $\sum_{k=0}^{20}\frac{3^k}{k!}\approx 20.085$
A: With the error term of the taylor series you just need to calculate 
$$\frac{x^n}{n!}< \frac{1}{100}$$
There is no biggest $n$ as it will be true for any bigger $n$ to as the factorial grows faster. You need a bound for $x$ to get an answer, for example if you calculate it on $[-1,1]$  you need $n=5$.
A: It depends on $x$.  Just keep adding terms until the term you add is less than $.01$.
For example, if $x=1$ you have $1/0! = 1 > .01 $, $1/1! = 1 > .01$, ..., $1/4! = 1/24 > .01$,  $1/5! = 1/120< .01$, so there you stop.
