Editing Proof that e is irrational (section)

==Fourier's proof==
The most well-known proof is [[Joseph Fourier]]'s [[proof by contradiction]],<ref>{{Cite book | last1 = de Stainville | first1 = Janot | year = 1815 | title = Mélanges d'Analyse Algébrique et de Géométrie |trans-title=A mixture of Algebraic Analysis and Geometry | publisher = Veuve Courcier | pages = 340–341}}</ref> which is based upon the equality

: <math>e = \sum_{n = 0}^\infty \frac{1}{n!}.</math>

Initially ''e'' is assumed to be a rational number of the form {{sfrac|''a''|''b''}}. The idea is to then analyze the scaled-up difference (here denoted ''x'') between the series representation of ''e'' and its strictly smaller {{nowrap|''b''-th}} partial sum, which approximates the limiting value ''e''. By choosing the scale factor to be the [[factorial]] of&nbsp;''b'', the fraction {{sfrac|''a''|''b''}} and the {{nowrap|''b''-th}} partial sum are turned into [[integer]]s, hence ''x'' must be a positive integer. However, the fast convergence of the series representation implies that ''x'' is still strictly smaller than&nbsp;1. From this contradiction we deduce that ''e'' is irrational.

Now for the details. If ''e'' is a [[rational number]], there exist positive integers ''a'' and ''b'' such that {{nowrap|1=''e'' = {{sfrac|''a''|''b''}}}}. Define the number

: <math>x = b!\left(e - \sum_{n = 0}^{b} \frac{1}{n!}\right).</math>

Use the assumption that ''e'' = {{sfrac|''a''|''b''}} to obtain

: <math>x = b!\left (\frac{a}{b} - \sum_{n = 0}^{b} \frac{1}{n!}\right) = a(b - 1)! - \sum_{n = 0}^{b} \frac{b!}{n!}.</math>

The first term is an integer, and every fraction in the sum is actually an integer because {{nowrap|''n'' ≤ ''b''}} for each term. Therefore, under the assumption that ''e'' is rational, ''x'' is an integer.

We now prove that {{nowrap|0 < ''x'' < 1}}. First, to prove that ''x'' is strictly positive, we insert the above series representation of ''e'' into the definition of ''x'' and obtain

: <math>x = b!\left(\sum_{n = 0}^{\infty} \frac{1}{n!} - \sum_{n = 0}^{b} \frac{1}{n!}\right) = \sum_{n = b+1}^{\infty} \frac{b!}{n!}>0,</math>

because all the terms are strictly positive.

We now prove that {{nowrap|''x'' < 1}}. For all terms with {{nowrap|''n'' ≥ ''b'' + 1}} we have the upper estimate

: <math>\frac{b!}{n!} =\frac1{(b + 1)(b + 2) \cdots \big(b + (n - b)\big)} \le \frac1{(b + 1)^{n-b}}.</math>

This inequality is strict for every {{nowrap|''n'' ≥ ''b'' + 2}}. Changing the index of summation to {{nowrap|1=''k'' = ''n'' – ''b''}} and using the formula for the [[Geometric series|infinite geometric series]], we obtain

:<math>x =\sum_{n = b + 1}^\infty \frac{b!}{n!}
< \sum_{n=b+1}^\infty \frac1{(b + 1)^{n-b}}
=\sum_{k=1}^\infty \frac1{(b + 1)^k}
=\frac{1}{b+1} \left (\frac1{1 - \frac1{b + 1}}\right)
= \frac{1}{b} \le 1.</math>

And therefore <math>x<1.</math>

Since there is no integer strictly between 0 and 1, we have reached a contradiction, and so ''e'' is irrational, [[Q.E.D.]]