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Theorem 1   For any algebraic number $\alpha$ with degree $m>1$ , there exists a constant $c=c(\alpha)>0$ such that: $$|\alpha-\frac{p}{q}|> \frac{c}{q^m}$$ for all rationals $p/q$ (with $q>0$ ).

Corollary 1   The real number $$\psi= \sum_{n=1}^\infty \frac{1}{10^{n!}}=0.1100010\ldots$$ is transcendental.

Proof. Clearly, the number $\psi$ is well defined, i.e. the series converges. Indeed, $$\frac{1}{10^{n!}}<\frac{1}{10^n}$$ and $\sum_{n=1}^\infty 10^{-n}=1/9$ . Thus, by the comparison test, the series converges and $0<\psi<1/9$ .

Suppose, for a contradiction, that $\psi$ is algebraic of degree $m$ . We will construct infinitely many rationals $p/q$ such that $$|\psi - \frac{p}{q}|< \frac{c}{q^m}$$ where $c=c(\psi)$ is the constant given by the theorem above. Let $k\in \Nats$ be such that $1/2^k < c$ . Then, in fact, we will show that there are infinitely many rationals $p/q$ with $q\geq 2$ such that $$|\psi -\frac{p}{q}|<\frac{1}{q^{m+k}}<\frac{1}{2^k}\cdot \frac{1}{q^m}<\frac{c}{q^m}$$ For all $j>k+m$ we define a rational number $p_j/q_j$ by: $$p_j=10^{j!}\sum_{n=1}^j 10^{-n!},\quad q_j=10^{j!}$$ then $p_j$ and $q_j$ are relatively prime integers and we have: \begin{eqnarray*} |\psi - \frac{p_j}{q_j}| & = & \sum_{n=j+1}^\infty \frac{1}{10^{n!}}\\ & < & \frac{1}{10^{(j+1)!}}(1+\frac{1}{10}+\frac{1}{10^2}+\ldots)\\ & = & 10/9\cdot \frac{1}{q_j^{(j+1)}} \\ & < & \frac{1}{q_j^j}\\ & < & \frac{1}{q_j^{(k+m)}} \end{eqnarray*}where in the last inequality we have used the fact that $j>k+m$ . Therefore, all rationals $\{ p_j/q_j \}_{j=k+m+1}^\infty$ satisfy the desired inequality, which leads to the contradiction with the theorem above. Thus $\psi$ cannot be algebraic and it must be transcendental.

example of transcendental number is owned by alozano.