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)

\begin{align} \quad \left | P \left ( \frac{a}{b} \right ) \right | \geq \frac{1}{b^d} \end{align}

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)

\begin{align} \quad \left | \alpha - \frac{a}{b} \right | \geq \frac{C(\alpha)}{b^d} \end{align}

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)

\begin{align} \quad \sum_{k=0}^{\infty} 10^{-k!} = \sum_{k=0}^{\infty} \frac{1}{10^{k!}} \leq \sum_{k=0}^{\infty} \frac{1}{2^n} = 2 < \infty \end{align}

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)

\begin{align} \quad \left | \alpha - \frac{a}{b} \right | \geq \frac{C(\alpha)}{b^d} \end{align}

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

(5)

\begin{align} \quad s_n = \sum_{k=0}^{n} 10^{-k!} \end{align}

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)

\begin{align} \quad |\alpha - s_n| \geq \frac{C(\alpha)}{(10^{n!})^d} \quad (*) \end{align}

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

(7)

\begin{align} \quad |\alpha - s_n| = \left | \sum_{k=n+1}^{\infty} 10^{-k!} \right | \leq \sum_{k=0}^{\infty} \frac{10^{-(n+1)!}}{10^k} = \frac{\left ( \frac{10}{9} \right )}{10^{(n+1)!}} \quad (**) \end{align}

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

(8)

\begin{align} \quad \frac{C(\alpha)}{(10^{n!})^d} \leq |\alpha - s_n| \leq \frac{\left ( \frac{10}{9} \right )}{10^{(n+1)!}} \end{align}

Hence for each $n \in \mathbb{N}$

(9)

\begin{align} \quad \frac{C(\alpha)}{\left (\frac{10}{9} \right )} \leq \frac{10^{n!d}}{10^{(n+1)!}} \end{align}

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.