Calculate the number e using Taylor polynomial and Lagrange Remainder I am supposed to determine the value of number e with exactness (mistake) less than $10^{-8}$, using Taylor polynomial and Lagrange Remainder. 
My hint is that $T_{n}(e^{x},0)(x)$.
I know the formula for Lagrange Remainder, as well as for Taylor polynomial of e, but I am not able to combine it.
Can anyone help me? 
 A: Let the $n$th Taylor polynomial of $e^x$ be $T_n$ with corresponding remainder $R_n$. To approximate the value of $e$, you want to focus on evaluating the Taylor polynomial at $x = 1$, i.e., $T_x(1)$ because $e^1 = e$.
Your goal is to find some positive integer $n$ such that $|T_n(1) - e^1| = |R_n(1)| < 10^{-8}$. You said you have a way to calculate the remainders, so the easiest way to solve your problem would be to evaluate $|R_n(1)|$ for each $n = 1, 2, 3...$ until you get a value less than $10^{-8}$.
Once you find a suitable $n$, evaluate $T_n(1)$ for that $n$ to get your approximation.
A: $$e=\sum_{k=0}^\infty \ \frac{1}{k!}=\sum_{k=0}^N\ \frac{1}{k!}+\sum_{k=N+1}^\infty\ \frac{1}{k!}=\sum_{k=0}^N\ \frac{1}{k!}+\frac{1}{(N+1)!}+\frac{1}{(N+2)!}+...$$
Note the remainder term is :
$$\frac{1}{(N)!}[\frac{1}{N+1}+\frac{1}{(N+1)(N+2)}+\frac{1}{(N+1)(N+2)(N+3)}+...]<\frac{1}{N!}[\frac{1}{(N+1)}+\frac{1}{(N+1)^2}+\frac{1}{(N+1)^3}+...]=\frac{1}{N\cdot N!}$$
So you can probably get away with $\frac{1}{(N+1)!}<10^{-8}$.
$N\approx 11. $
A: The Lagrange remainder can be expressed as
$$\frac{t^{n+1}e^t}{(n+1)!}$$ where $0<t<1$ and is bounded by
$$\frac1{(n+1)!}.$$
Then use $$11!<10^8<12!$$

$$\sum_{n=0}^{11}\dfrac1{n!}=\frac{13563139}{4989600}.$$

