How can the value of $e$ be found to five-place accuracy using Higher Order Mean Value Theorem. Hint: Use the fact that $e\approx 3$. [closed]

Instead, we shall use $e<3.$

$$e=\sum_{n=0}^{\infty}\frac{1}{n!}=\sum_{n=0}^m \frac{1}{n!} + \sum_{n=m+1}^\infty \frac{1}{n!}$$

$$ \sum_{n=m+1}^\infty \frac{1}{n!} < \frac{1}{(m+1)!}(\sum_{n=0}^\infty\frac{1}{n!})=\frac{1}{(m+1)!}e<\frac{3}{(m+1)!}$$

Therefore, we have $0<e-\sum_{n=1}^m\frac{1}{n!}<\frac{3}{(m+1)!}$ for all $m$.

When $m=8$, $\frac{3}{9!}<10^{-5}$, hence we know that $e$ agrees with $\sum_{n=1}^8\frac{1}{n!}=2.71827...$ at least up to $2.7182$.

Just for your curiosity.

Starting from @Just a user's answer, suppose that you want to know $m$ such that $$\frac{3}{(m+1)!} \leq \epsilon \implies (m+1)!~ \geq ~\frac 3 \epsilon$$ ANswering this question @Gary provided a superb approximation of the inverse of the factorial function in terms of Lambert function.

Applied to your case, $$m \sim \frac{{\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)}}{{W\left( {\frac{1}{e}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)} \right)}} - \frac{3}{2} \qquad \text{where}\qquad y=\frac 3 \epsilon$$ Suppose that we want $\epsilon=10^{-20}$, this would give as a real $m=20.5739$ that is to say $\lceil m\rceil=21$.