Calculate the value of $e$ from integral definition Starting with the definition of $e$ as $$\int_1^e \frac{dx}{x} = 1,$$ how can I show that $e = 2.718\ldots$?
 A: Hint. An elementary approach. One may consider
$$
I_0=1-\frac1{e}, \quad I_n=\int_{\large\frac1{e}}^1\left(-\ln x \right)^n\:dx, \quad n\ge1,
$$ then integrating by parts,
$$
\begin{align}
I_n&=\left[ x \frac{}{}\left(-\ln x \right)^n\right]_{\large\frac1{e}}^1 +n\int_{\large\frac1{e}}^1\left(-\ln x \right)^{n-1}\:dx
\\&=-\frac1{e}+nI_{n-1}
\end{align}
$$ dividing by $n!$ gives
$$
\frac{I_n}{n!}-\frac{I_{n-1}}{(n-1)!}=-\frac1{e}\cdot \frac1{n!},\qquad n\ge1,
$$ by summing, terms telescope to get
$$
e=1+\frac1{1!}+\cdots+\frac1{n!}+\frac1{n!}\int_{\large\frac1{e}}^1\left(-\ln x \right)^n\:dx,
$$ then one may observe that
$$
0<\int_{\large\frac1{e}}^1\left(-\ln x \right)^n\:dx< 2
$$ giving, for $n=1,2,\cdots$,

$$
1+\frac1{1!}+\cdots+\frac1{n!}<e<1+\frac1{1!}+\cdots+\frac1{n!}+\frac2{n!}
$$ 

which may be useful for a numerical approximation of $e$.
A: Though it is a bit roundabout, you can define $\ln(x)=\int_1^x \frac{1}{t} dt$ and $\exp$ to be the inverse of $\ln$. It then follows that $\exp(1)$ is your $e$. You can now develop the Taylor series of $\exp$. From this development you find $e=\sum_{n=0}^\infty \frac{1}{n!}$ and also 
$$0<e-\sum_{n=0}^N \frac{1}{n!}<\frac{e}{(N+1)!}.$$ 
This follows from the Lagrange error formula and a little bit of estimation. Hence we have
$$\sum_{n=0}^N \frac{1}{n!}<e<\frac{(N+1)!}{(N+1)!-1} \sum_{n=0}^N \frac{1}{n!}.$$
You wanted to infer the first four digits, which can be done with $N=6$: rounding the lower bound down to five digits and the upper bound up to five digits, you get $2.7180<e<2.7186$.
You can also attempt to directly manipulate the integral numerically; for example, for any $b \geq 1$ and positive integer $N$, you have
$$\sum_{i=1}^N \frac{1}{1+\frac{b-1}{N} i} \frac{b-1}{N} \leq \int_1^b \frac{1}{x} dx \leq \sum_{i=0}^{N-1} \frac{1}{1+\frac{b-1}{N} i} \frac{b-1}{N}.$$
So to prove that $e>b$, it is enough to prove that for some $N$, $\sum_{i=0}^{N-1} \frac{1}{1+\frac{b-1}{N} i} \frac{b-1}{N}<1$. Similarly to prove $e<b$ it is enough to prove that for some $N$, $\sum_{i=1}^N \frac{1}{1+\frac{b-1}{N} i} \frac{b-1}{N}>1$. This will be much more computationally expensive than the previous approach, however.
