Calculate the value of $e$ from integral definition

Question

Starting with the definition of $e$ as $$\int_1^e \frac{dx}{x} = 1,$$ how can I show that $e = 2.718\ldots$?

Answer 1

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 $e$.

Answer 2

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.