Exercise 6.2.5 ($\sum_{j=1}^\infty 2^{-3^j}, \ \sum_{j=1}^\infty 2^{-j!}$ are irrational)

Proof.

(a)

Put $$\beta =\underset{j=1}{\overset{\infty }{\sum }}{2}^{-3^j},{\beta }_n=\underset{j=1}{\overset{n}{\sum }}{2}^{-3^j}.$$

(Since $3^j\ge 2^j>j$, we obtain $0<2^{-3^j}<2^{-j}$, where ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^{\infty }{2}^{-j}$ is convergent, thus ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^{\infty }{2}^{-3^j}$ converges.)

Then

$${\beta }_n=\frac{h_n}{k_n}\in \mathbb{Q},\text{where }\{\begin{array}{c}{h}_n=2^{3^n-3}+2^{3^n-3^2}+\cdots +1,\hfill \\ k_n=2^{3^n}.\hfill \end{array}$$

Since the sequence ${({\beta }_n)}_{n\in {\mathbb{N}}^{\text{∗}}}$ is strictly increasing, all the ${\beta }_n=h_n∕k_n$ are distinct rational numbers. Moreover

$$\begin{array}{llllll}\hfill 0\le \beta -\frac{h_n}{k_n}& =\underset{j=n+1}{\overset{\infty }{\sum }}{2}^{-3^j}& \hfill & & \hfill & \\ \hfill & =2^{-3^{n+1}}\underset{i=1}{\overset{\infty }{\sum }}\frac{1}{2^{3^{n+i}-3^{n+1}}}& \hfill (j=n+i)& & \hfill & & \hfill \\ \hfill & =2^{-3^{n+1}}\underset{i=1}{\overset{\infty }{\sum }}\frac{1}{2^{3^{n+1}(3^{i-1}-1)}}.& \hfill & & \hfill & \end{array}$$

We know that for all $n\in \mathbb{N}$, $2^n>n$, so $2^n\ge n+1$. Then $3^{i-1}-1\ge 2^{i-1}-1\ge i-1$ if $i\ge 1$. Therefore

$$\begin{array}{llllll}\hfill 0\le \beta -\frac{h_n}{k_n}& \le 2^{-3^{n+1}}\underset{i=1}{\overset{\infty }{\sum }}\frac{1}{2^{(i-1)3^{n+1}}}& \hfill & & \hfill & \\ \hfill & =2^{-3^{n+1}}\underset{i=1}{\overset{\infty }{\sum }}\frac{1}{a^{i-1}}& \hfill (\text{where }a=2^{3^{n+1}}=k_n^3)& & \hfill & & \hfill \\ \hfill & =\frac{1}{a}\frac{1}{1-\frac{1}{a}}& \hfill & & \hfill & \\ \hfill & =\frac{1}{k_n^3-1}.& \hfill & & \hfill & \end{array}$$

Note that

$$\frac{1}{k_n^3-1}\le \frac{1}{k_n^2}⟺1\le k_n^2(k_n-1),$$

so this inequality is true for all $n\in {\mathbb{N}}^{\text{∗}}$. This shows that for all $n\in {\mathbb{N}}^{\text{∗}}$

$$\left|\beta -\frac{h_n}{k_n}\right|\le \frac{1}{k_n^2}.$$

So there are infinitely many rationals $h∕k$ such that

$$\left|\beta -\frac{h}{k}\right|\le \frac{1}{k^2}.$$

By Problem 4, $\beta ={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^{\infty }{2}^{-3^j}$ is irrational.

(b)

Put $$\gamma =\underset{j=1}{\overset{\infty }{\sum }}{2}^{-j!},{\gamma }_n=\underset{j=1}{\overset{n}{\sum }}{2}^{-j!}.$$

(As in part (a), $j!\ge j$, so ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^{\infty }{2}^{-j!}$ converges.) Then

$${\gamma }_n=\frac{h_n}{k_n}\in \mathbb{Q},\text{where }\{\begin{array}{c}{h}_n=2^{n!-1!}+2^{n!-2!}+\cdots +1,\hfill \\ k_n=2^{n!}.\hfill \end{array}$$

Since the sequence ${({\gamma }_n)}_{n\in {\mathbb{N}}^{\text{∗}}}$ is strictly increasing, all the ${\gamma }_n=h_n∕k_n$ are distinct rational numbers. Moreover

$$\begin{array}{llll}\hfill 0\le \gamma -\frac{h_n}{k_n}& =\underset{j=n+1}{\overset{\infty }{\sum }}{2}^{-j!}& \hfill & \\ \hfill & =\frac{1}{2^{(n+1)!}}\underset{i=0}{\overset{\infty }{\sum }}\frac{1}{2^{(n+1-i)!-(n+1)!}}.& \hfill & \end{array}$$

For all $i\in \mathbb{N}$,

$$\begin{array}{llll}\hfill (n+1-i)!-(n+1)!& =(n+1)!\left[\frac{(n+1+i)!}{(n+1)!}-1\right]& \hfill & \\ \hfill & =(n+1)!\left[(n+1+i)(n+i)\cdots (n+2)-1\right]& \hfill & \\ \hfill & \ge (n+1)![{(n+1)}^i-1]& \hfill & \\ \hfill & \ge (n+1)![2^i-1]& \hfill & \\ \hfill & \ge (n+1)!i.& \hfill & \end{array}$$

Therefore

$$\begin{array}{llllll}\hfill 0\le \gamma -\frac{h_n}{k_n}& \le \frac{1}{2^{(n+1)!}}\underset{i=0}{\overset{\infty }{\sum }}\frac{1}{2^{i(n+1)!}}& \hfill & & \hfill & \\ \hfill & =\frac{1}{2^{(n+1)!}-1}& \hfill & & \hfill & \\ \hfill & \le \frac{2}{2^{(n+1)!}}& \hfill & & \hfill & \\ \hfill & =\frac{2}{k_n^{n+1}}& \hfill & & \hfill & \\ \hfill & \le \frac{2}{k_n^2}& \hfill (\text{for }n\ge 1).& & \hfill & & \hfill \end{array}$$

So there are infinitely many rationals $h∕k$ such that

$$\left|\gamma -\frac{h}{k}\right|\le \frac{2}{k^2}.$$

By Problem 4, $\gamma ={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^{\infty }{2}^{-j!}$ is irrational.