# An Example of a Series Converging to a Transcendental Number

Recall from the Algebraic Number Approximation page that if $P \in \mathbb{Z}[x]$ is a polynomial of degree $d$ and $\frac{a}{b} \in \mathbb{Q}$ is such that $P \left ( \frac{a}{b} \right ) \neq 0$ then:

(1)We then proved that if $\alpha$ is an algebraic number of degree $d$ then there exists a number $C(\alpha) > 0$ such that for all $\frac{a}{b} \in \mathbb{Q}$ with $\alpha \neq \frac{a}{b}$ we have that:

(2)We will now use the results above to show a particular series converges to a transcendental number.

## Example

**Show that the series $\displaystyle{\sum_{k=0}^{\infty} 10^{-k!}}$ converges to a transcendental number.**

First observe that the series does indeed converge since:

(3)Let $\alpha = \sum_{k=0}^{\infty} 10^{-k!}$. Suppose that there exists a polynomial $P \in \mathbb{Z}[x]$ of minimal degree $d$ for which $P(\alpha) = 0$. Then there exists a $C(\alpha) > 0$ such that for all $\frac{a}{b} \in \mathbb{Q}$ where $\alpha \neq \frac{a}{b}$ we have that:

(4)For each $n \in \mathbb{N}$ let $s_n$ denote the $n^{\mathrm{th}}$ partial sum of the series, that is:

(5)Each $s_n \in \mathbb{Q}$ and has denominator $b = 10^{n!}$. For each $n \in \mathbb{N}$ we have that $s_n \neq \alpha$ and so:

(6)Also, for each $n \in \mathbb{N}$ we have that:

(7)So from $(*)$ and $(**)$ we have that:

(8)Hence for each $n \in \mathbb{N}$

(9)As $n \to \infty$ the righthand side goes to $0$ which is a contradiction since $C(\alpha) > 0$. Hence the assumption that $\alpha$ is algebraic is false. So $\alpha$ is transcendental.