Expression for $e^1$

Question

I am currently looking at "Calculus" by Michael Spivak and in his proof for the irrationality of $e$ he writes the following

"We know that, for any n,

$e=e^1=1+\frac{1}{1!}+\frac{1}{2!}+\cdots+\frac{1}{n!}+R_n$ where $0<R_n<\frac{3}{(n+1)!}$"

I was just wondering where this $R_n$ came from, why is the numerator 3?

Thanks in advance.

Answer 1

We can start with the remainder term of Taylor's theorem. In the case where we approximate $e^1$ by a Taylor series for $e^x$ centered at $0$, we have $$ e = 1 + \frac1{1!} + \frac{1}{2!} + \frac1{3!} + \dots + \frac1{n!} + \frac{e^\xi}{(n+1)!} $$ for some $\xi \in [0,1]$. In particular, since $e^x$ is increasing from $0$ to $1$, we know that the remainder term satisfies $R_n \le \frac{e^1}{(n+1)!}$.

But this isn't a very useful formula for the error term, since we're trying to figure out what $e$ is. So let's take a small instance of this formula and use it to get an upper bound on $e$. For example, we can take $$ e = 1 + \frac1{1!} + \frac1{2!} + \frac{e^{\xi}}{3!} \le 1 + 1 + \frac12 + \frac e6 \implies \frac56 e \le \frac52 \implies e \le 3. $$ Now we can say that the error term satisfies $R_n \le \frac{e}{(n+1)!} \le \frac{3}{(n+1)!}$.

Answer 2

Write Lagrange's remainder: $$R_n = \frac{f^{(n+1)}(\xi)}{(n+1)!}(b-a)^{n+1}$$ for $\xi \in (a,b)$.

Since we are on $(0,1)$ and $f^{(n+1)}(x) = e^x$, you get: $$R_n = \frac{e^{\xi}}{(n+1)!} < \frac{3}{(n+1)!}$$ being $\xi \in (0,1)$ and, thus, $e^{\xi} < 3$.

Answer 3

The tail of the sum is

$$\sum_{k={n+1}}^\infty\frac1{k!}=\frac1{(n+1)!}\sum_{k=0}^\infty\frac{(n+1)!}{(n+1+k)!}<\frac1{(n+1)!}\sum_{k=0}^\infty\frac{1}{k!}=\frac e{(n+1)!}.$$

The constant could be made tighter, but this is not crucial.