Least integer $n$ such that $|exp(1)-S_n| \le 10^{-5}$

Question

The problem asks to determine the least natural number $n$ such that

$$\left| e-\displaystyle\sum_{k=0}^n \dfrac{1}{k!} \right| \le 10^{-5}$$

My approach

By definition of $e$

$$e-\displaystyle\sum_{k=0}^n \dfrac{1}{k!} = \displaystyle\sum_{k=n+1}^{+\infty} \dfrac{1}{k!} $$

One may write the RHS as

$$\displaystyle\sum_{k=n+1}^{+\infty} \dfrac{1}{k!} = \dfrac{1}{(n+1)!} \left(1 + \dfrac{1}{n+2} + \dfrac{1}{(n+2)(n+3)}+ \cdots \right) $$

Then bound above the sum between brackers by the geometric series sum

$$\displaystyle\sum_{i=0}^{+\infty} \left(\dfrac{1}{n+2}\right)^i$$

We get

$$\left| e-\displaystyle\sum_{k=0}^n \dfrac{1}{k!} \right| \le \dfrac{(n+2)}{(n+1).(n+1)!}$$

For the LHS to be $\le 10^{-5}$, it suffices (only :-/) to have $\dfrac{(n+1).(n+1)!}{n+2} \ge 10^5$ which holds for $n = 9$ (calculations) and greater values of $n$ since left side increases with $n$.

How to prove that $9$ is the least such integer.

Thanks.

Answer 1

You have done a good job showing that $$R_n=\left| e-\displaystyle\sum_{k=0}^n \dfrac{1}{k!} \right| \le \dfrac{(n+2)}{(n+1).(n+1)!}$$ Now, you search for $n$ such that $R_n < 10^{-k}$ that is to say $$\dfrac{(n+1).(n+1)!}{(n+2)}=10^k$$ which is "almost" $$(n+1)!=10^k$$ For the time being, let $m=n+1$ and have a look at the magnificent approximation @robjohn gave in this question of mine asking for the solution of $$m!=a^m 10^k$$ $$m\sim ea\exp\left(\operatorname{W}\left(\frac k{ea}\log(10)-\frac1{2ea}\log(2\pi a)\right)\right)-\frac12$$ where appears Lambert function.

Applied to your case $(a=1)$ this will give $$n=e \exp\left(\operatorname{W}\left(\frac{2k \log (10)-\log (2 \pi )}{2 e}\right)\right)-\frac32$$ For $k=5$, this gives $n=7.42$ and $\lceil n \rceil=8$.

Checking $R_7=2.79\times 10^{-5}$, $R_8=3.06\times 10^{-6}$.