# Liouville's Constant is Transcendental/Corollary

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## Corollary to Liouville's Constant is Transcendental

All real numbers of the form:

 $\ds \sum_{n \mathop = 1}^\infty \frac {a_n} {10^{n!} }$ $=$ $\ds \frac {a_1} {10^1} + \frac {a_2} {10^2} + \frac {a_3} {10^6} + \frac {a_4} {10^{24} } + \cdots$

where

$a_1, a_2, a_3, \ldots \in \set {1, 2, \ldots, 9}$

are transcendental.

## Proof

Let $n \in \N$.

For $n = 1$, let $p = a_1$ and $q = 10$. Then:

$\ds \size {L - \dfrac p q} = \sum_{k \mathop = 2}^\infty \dfrac {a_k} {10^{k!} } < \dfrac 1 {10} = \dfrac 1 q$

For $n > 1$, let $q = 10^{n!}$ and write:

$\ds L = \frac p q + \sum_{k \mathop = n + 1}^\infty \frac {a_k} {10^{k!} }$

for some suitable $p \in \Z$.

Then:

 $\ds \size {L - \frac p q}$ $=$ $\ds \sum_{k \mathop = n + 1}^\infty \frac {a_k} {10^{k!} }$ $\ds$ $\le$ $\ds \sum_{k \mathop = n + 1}^\infty \frac 9 {10^{k!} }$ $\ds$ $\le$ $\ds \frac {18} {10^{\paren {n + 1}!} }$ $\ds$ $=$ $\ds \frac {18} {q^{n + 1} }$ $\ds$ $<$ $\ds \frac 1 {q^n}$ as $q \ge 100$ for all $n > 1$

Thus, by definition, $L$ is a Liouville number.

Therefore, by Liouville's Theorem, $L$ is transcendental.

$\blacksquare$