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The Liouville approximation theorem states that an irrational algebraic number $\alpha$ of degree $d\ge 2$ satisfies $|\alpha -a/b|\ge C_{\alpha }{b}^{-d}$ for all integers $a$ and $b\ge 1$, for some $C_{\alpha }>0$. The constant is independent of the rational approximation.

The required condition is infinitely many of the coefficients equal one. If only finitely many do, the series is a finite sum of rational numbers. Conversely let $\alpha$ be the sum with infinitely many nonzero terms. Its truncation $s_N$ has denominator dividing $q_N=10^{N!}$. For $N\ge 2$, its strictly positive tail satisfiesIf $\alpha =a/b$ were rational, the distinct fractions $\alpha ,s_N$ would have distance at least $1/(b{q}_N)$, contradicting the tail bound for large $N$. Thus $\alpha$ is irrational. If it were algebraic of fixed degree $d$, apply the Liouville approximation theorem to $s_N$. Even if its reduced denominator is smaller than $q_N$, the lower bound is at least $C_{\alpha }{q}_N^{-d}$. The displayed upper bound contradicts this as $N\to \infty$. HenceThe coincidence $0!=1!$ changes only the first rational contribution and has no effect on this criterion.

Liouville approximation theorem states that if $\alpha$ is a real algebraic number of degree $d\ge 2$, then there is a constant $c(\alpha )>0$ such thatfor every rational $p/q$ with $q>0$. The strict inequality may be replaced by a non-strict one after changing the constant.

LetThe tail satisfiesFor every fixed positive integer $d$, this is smaller than $q_N^{-d}$ for all sufficiently large $N$.

The binary expansion has $1$ digits at factorial positions and arbitrarily long blocks of zeros, but it is neither eventually zero nor eventually periodic. Since a rational number has an eventually periodic binary expansion, $\alpha$ is irrational. If it were an algebraic number of degree $d$, part (b), or equivalently the Liouville approximation theorem, would give a fixed positive multiple of $q^{-d}$ as a lower bound for all rational approximations. The fractions $p_N/q_N$ violate that bound. Thus $\alpha$ is a transcendental number.

This is always true for every positive integer $n$. If ${\beta }^n$ were algebraic, then $\beta$ would be a root of the polynomial $X^n-{\beta }^n$ over the field of algebraic numbers. By transitivity of algebraic extensions, $\beta$ would then be algebraic over $\mathbb{Q}$, contradicting its transcendence.

This can be false. For any nonzero transcendental number $\beta$, the number $\gamma =1/\beta$ is also transcendental, since otherwise its reciprocal would be algebraic. Yet $\beta \gamma =1$ is algebraic.

LetIf $x$ were rational, thenwould make $\sqrt[3]{6}=(x^3-5)/(3x)$ rational. But if $\sqrt[3]{6}=a/b$ in lowest terms, then $a^3=6b^3$, which is impossible by unique prime factorization: the exponent of $2$ on the two sides is respectively a multiple of three and one more than a multiple of three. Therefore

Use the given convergent seriesSuppose this number were $a/b\in \mathbb{Q}$. Choose $N$ large enough that $b\mid (2N+1)!$. Multiplication by $(2N+1)!$ makes both the rational number and the partial sum through $n=N$ integers. Their differencewould therefore be an integer. It is positive, whilefor sufficiently large $N$, a contradiction. Hence

A transcendental number is a complex number that is not a root of any nonzero polynomial with rational, equivalently integer, coefficients. LetIf $a\ne 0$ and $y$ were an algebraic number, then $e$ would satisfyover the algebraic extension $\mathbb{Q}(y)$. By transitivity of algebraic extensions, $e$ would be algebraic over $\mathbb{Q}$, contradicting its transcendence. If $a=0$, then $b\ne 0$ and $e=b/y$ would again be algebraic. Thus

For the general result, choose logarithms of nonzero algebraic numbers ${\alpha }_1,\dots ,{\alpha }_s$ and algebraic coefficients ${\beta }_0,\dots ,{\beta }_s$, and putLetwhere $f_{{\alpha }_j}$ is the minimal polynomial and its $H$ is the naive polynomial height, and letThe general lower bound for a linear form in logarithms states that, if $\Lambda \ne 0$, thenwhere the effective constant $C$ depends only on $s$ and the degree of the number field generated by all the data.

For the improved homogeneous result, take $b_1,\dots ,b_s\in \mathbb{Z}$ andWith the same $A_j$, after ordering the terms setThe Baker lower bound for a homogeneous linear form in logarithms givesBoth constants are effective. The division by $\log A_s$ in $B^{\ast }$ is the improvement that matters when ${\alpha }_s$ has variable height.