Editing Liouville number (section)

=== Notes on the proof ===
# The inequality <math>\sum_{k=n+1}^\infty \frac{a_k}{b^{k!}} \le \sum_{k=n+1}^\infty \frac{b-1}{b^{k!}}</math> follows since ''a''<sub>''k''</sub>&nbsp;∈&nbsp;{0,&nbsp;1,&nbsp;2,&nbsp;..., ''b''−1} for all ''k'', so at most ''a''<sub>''k''</sub>&nbsp;= ''b''−1. The largest possible sum would occur if the sequence of integers (''a''<sub>1</sub>, ''a''<sub>2</sub>,&nbsp;...) were (''b''−1, ''b''−1, ...), i.e. ''a''<sub>''k''</sub>&nbsp;= ''b''−1, for all ''k''. <math>\sum_{k=n+1}^\infty \frac{a_k}{b^{k!}}</math> will thus be less than or equal to this largest possible sum.
# The strong inequality <math>\begin{align}
\sum_{k=n+1}^\infty \frac{b-1}{b^{k!}} < \sum_{k=(n+1)!}^\infty \frac{b-1}{b^k}
\end{align}</math> follows from the motivation to eliminate the [[series (mathematics)|series]] by way of reducing it to a series for which a formula is known. In the proof so far, the purpose for introducing the inequality in #1 comes from intuition that <math>\sum_{k=0}^\infty \frac{1}{b^{k}} = \frac{b}{b-1}</math> (the [[geometric series]] formula); therefore, if an inequality can be found from <math>\sum_{k=n+1}^\infty \frac{a_k}{b^{k!}}</math> that introduces a series with (''b''−1) in the numerator, and if the denominator term can be further reduced from <math>b^{k!}</math>to <math>b^{k}</math>, as well as shifting the series indices from 0 to <math>\infty</math>, then both series and (''b''−1) terms will be eliminated, getting closer to a fraction of the form <math>\frac{1}{b^{\text{exponent}\times n}}</math>, which is the end-goal of the proof. This motivation is increased here by selecting now from the sum <math>\sum_{k=n+1}^\infty \frac{b-1}{b^{k!}}</math> a partial sum. Observe that, for any term in <math>\sum_{k=n+1}^\infty \frac{b-1}{b^{k!}}</math>, since ''b'' ≥ 2, then <math>\frac{b-1}{b^{k!}} < \frac{b-1}{b^{k}}</math>, for all ''k'' (except for when ''n''=1). Therefore, <math>\begin{align}
\sum_{k=n+1}^\infty \frac{b-1}{b^{k!}} < \sum_{k=n+1}^\infty \frac{b-1}{b^k}
\end{align}</math> (since, even if ''n''=1, all subsequent terms are smaller). In order to manipulate the indices so that ''k'' starts at 0, partial sum will be selected from within <math>
\sum_{k=n+1}^\infty \frac{b-1}{b^k}
</math> (also less than the total value since it is a partial sum from a series whose terms are all positive). Choose the partial sum formed by starting at ''k'' = (''n''+1)! which follows from the motivation to write a new series with ''k''=0, namely by noticing that <math>b^{(n+1)!} = b^{(n+1)!}b^0</math>.
#For the final inequality <math>\frac{b}{b^{(n+1)!}} \le \frac{b^{n!}}{b^{(n+1)!}}</math>, this particular inequality has been chosen (true because ''b'' ≥ 2, where equality follows [[if and only if]] ''n''=1) because of the wish to manipulate <math>\frac{b}{b^{(n+1)!}}</math> into something of the form <math>\frac{1}{b^{\text{exponent}\times n}}</math>. This particular inequality allows the elimination of (''n''+1)! and the numerator, using the property that (''n''+1)! – ''n''! = (''n''!)''n'', thus putting the denominator in ideal form for the substitution <math>q_n = b^{n!}</math>.