Let $a$ and $b$ be the lower and upper bound of $e$, respectively. Both $a$ and $b$ are rational numbers. Without using a calculator and without knowing the value of $e$, find $a$ and $b$ where $b-a<10^{-6}$.
So, we are asked to find
$$a<e<b$$
where $b-a<10^{-6}$. Using the series expansion for $e$, I'm able to find both $a$ and $b$, namely
$$a=\sum_{n=0}^{10}\frac{1}{n!}=\frac{6235301}{3628800}$$
and
$$b=\sum_{n=0}^{9}\frac{1}{n!}=\frac{62353}{36288}$$
but I have to confirm my calculation for $n=11$ and $n=12$ so that the value of summation is more stable. We then have
$$b-a=\frac{1}{3628800}<10^{-6}$$
I also found another way using the definition of $e$ to obtain other values of $a$ and $b$ by using this inequality:
$$\left(1+\frac{1}{x}\right)^x < e < \left(1+\frac{1}{x}\right)^{x+1}$$
Though it's doable, this method is very time consuming and very tedious. Is there another way? If any, an efficient one?
The "tail" when we truncate the usual series for $e$ just after the term $\frac{1}{9!}$ is $$\frac{1}{10!}+\frac{1}{11!}+\frac{1}{12!}+\cdots.$$ This sum is less than the sum of the infinite geometric series $$\frac{1}{10!}\left(1+\frac{1}{11}+\frac{1}{11^2}+\cdots\right),$$ which is $\frac{11}{10\cdot 10!}$. This is less than our target error.
So we can take $a=\sum_0^9 \frac{1}{k!}$ and $b=a+\frac{11}{10\cdot 10!}$.