[proofplan] We argue by contradiction: assume that the Liouville number $L$ is algebraic, and let $p \in \mathbb{Z}[x]$ be a non-zero [polynomial](/pages/1305) with $p(L)=0$. The rational partial sums $L_n$ have denominator $10^{n!}$, so $p(L_n)$ is either zero or has absolute value at least $10^{-dn!}$, where $d=\deg p$. On the other hand, the tail estimate for $L-L_n$ and a Lipschitz bound for $p$ on $[0,1]$ force $|p(L_n)|$ to be at most a constant times $10^{-(n+1)!}$, which is eventually smaller than $10^{-dn!}$. This contradiction proves that $L$ is not algebraic. [/proofplan] [step:Assume algebraicity and introduce the rational approximants] Suppose, for contradiction, that $L$ is algebraic. Then there exists a non-zero polynomial $p \in \mathbb{Z}[x]$ such that $p(L)=0$. Write \begin{align*} p(x) = a_d x^d + a_{d-1}x^{d-1} + \cdots + a_1x + a_0, \end{align*} where $d \in \mathbb{N}$, $a_d \ne 0$, and $a_0,\dots,a_d \in \mathbb{Z}$. For each $n \in \mathbb{N}$, define the $n$th partial sum \begin{align*} L_n := \sum_{k=1}^{n} \frac{1}{10^{k!}}. \end{align*} Since every term in the [series](/pages/1364) defining $L$ is positive, we have $0<L_n<L$ for every $n \in \mathbb{N}$. Moreover, because $k! \ge k$ for every $k \in \mathbb{N}$, each term satisfies $10^{-k!} \le 10^{-k}$. Therefore the geometric-series bound gives \begin{align*} L = \sum_{k=1}^{\infty}\frac{1}{10^{k!}} \le \sum_{k=1}^{\infty}\frac{1}{10^k} = \frac{1}{9} <1. \end{align*} Thus $0<L_n<L<1$ for every $n \in \mathbb{N}$, and in particular $L_n,L \in [0,1]$. [/step] [step:Bound the change of $p$ on the interval $[0,1]$] Define the constant \begin{align*} K := d\sum_{j=1}^{d} |a_j|. \end{align*} For all $x,y \in [0,1]$, the factor identity \begin{align*} x^j-y^j = (x-y)\sum_{m=0}^{j-1} x^{j-1-m}y^m \end{align*} gives \begin{align*} |p(x)-p(y)| &= \left|\sum_{j=1}^{d} a_j(x^j-y^j)\right| \\ &\le |x-y|\sum_{j=1}^{d}|a_j|\sum_{m=0}^{j-1}|x|^{j-1-m}|y|^m \\ &\le |x-y|\sum_{j=1}^{d}j|a_j| \\ &\le K|x-y|. \end{align*} [guided] We need a uniform estimate for how much $p$ can change when its input changes inside $[0,1]$. Recall that the polynomial is \begin{align*} p(x) = \sum_{j=0}^{d} a_jx^j, \end{align*} where $d \in \mathbb{N}$, $a_d \ne 0$, and $a_0,\dots,a_d \in \mathbb{Z}$. Define \begin{align*} K := d\sum_{j=1}^{d} |a_j|. \end{align*} This constant depends only on the integer coefficients of $p$ and its degree $d$. Let $x,y \in [0,1]$. For each integer $j$ with $1 \le j \le d$, we use the algebraic factorisation \begin{align*} x^j-y^j = (x-y)\sum_{m=0}^{j-1} x^{j-1-m}y^m. \end{align*} Substituting this identity into the polynomial difference gives \begin{align*} p(x)-p(y) = \sum_{j=1}^{d} a_j(x^j-y^j) = (x-y)\sum_{j=1}^{d}a_j\sum_{m=0}^{j-1}x^{j-1-m}y^m. \end{align*} Taking absolute values and applying the triangle inequality, \begin{align*} |p(x)-p(y)| &\le |x-y|\sum_{j=1}^{d}|a_j|\sum_{m=0}^{j-1}|x|^{j-1-m}|y|^m. \end{align*} Because $x,y \in [0,1]$, every factor $|x|^{j-1-m}|y|^m$ is at most $1$. Hence \begin{align*} |p(x)-p(y)| &\le |x-y|\sum_{j=1}^{d}j|a_j| \\ &\le |x-y|d\sum_{j=1}^{d}|a_j| \\ &= K|x-y|. \end{align*} Thus $p$ is Lipschitz on $[0,1]$ with Lipschitz constant $K$. [/guided] [/step] [step:Estimate the tail of the Liouville series] For every $n \in \mathbb{N}$, \begin{align*} |L-L_n| &= \sum_{k=n+1}^{\infty}\frac{1}{10^{k!}} \\ &= \frac{1}{10^{(n+1)!}}\left(1+\sum_{k=n+2}^{\infty}\frac{1}{10^{k!-(n+1)!}}\right). \end{align*} For $k \ge n+2$, the exponent $k!-(n+1)!$ is a positive integer, and the corresponding tail is bounded by the full geometric tail in base $10$. Therefore \begin{align*} |L-L_n| &\le \frac{1}{10^{(n+1)!}}\sum_{j=0}^{\infty}\frac{1}{10^j} \\ &= \frac{1}{10^{(n+1)!}}\cdot \frac{10}{9} \\ &\le \frac{2}{10^{(n+1)!}}. \end{align*} [/step] [step:Show that non-zero values of $p(L_n)$ have a denominator-controlled lower bound] For each $n \in \mathbb{N}$, define the integer \begin{align*} s_n := \sum_{k=1}^{n}10^{n!-k!}. \end{align*} Then \begin{align*} L_n = \frac{s_n}{10^{n!}}. \end{align*} Evaluating $p$ at $L_n$ and putting all terms over the common denominator $10^{dn!}$ gives \begin{align*} p(L_n) &= \sum_{j=0}^{d}a_j\left(\frac{s_n}{10^{n!}}\right)^j \\ &= \frac{1}{10^{dn!}}\sum_{j=0}^{d}a_js_n^j10^{(d-j)n!}. \end{align*} Define \begin{align*} t_n := \sum_{j=0}^{d}a_js_n^j10^{(d-j)n!} \in \mathbb{Z}. \end{align*} Thus $p(L_n)=t_n/10^{dn!}$. By the [Factor Theorem](/theorems/3235), each root $r$ of $p$ gives a linear factor $x-r$ of $p$. Iterating this factorisation over distinct roots shows that a non-zero polynomial of degree $d$ has at most $d$ roots. The sequence $(L_n)_{n=1}^{\infty}$ is strictly increasing because \begin{align*} L_{n+1}-L_n = \frac{1}{10^{(n+1)!}}>0, \end{align*} so it contains no repeated values. Hence only finitely many of the numbers $L_n$ can be roots of $p$. Therefore there exists $N_0 \in \mathbb{N}$ such that $p(L_n)\ne 0$ for all $n \ge N_0$. For $n \ge N_0$, we have $t_n \ne 0$, so $|t_n| \ge 1$. Hence \begin{align*} |p(L_n)| = \frac{|t_n|}{10^{dn!}} \ge \frac{1}{10^{dn!}}. \end{align*} [guided] The goal is to use the special decimal form of $L_n$ to obtain a lower bound for $|p(L_n)|$ whenever $p(L_n)$ is non-zero. For each $n \in \mathbb{N}$, define \begin{align*} s_n := \sum_{k=1}^{n}10^{n!-k!}. \end{align*} Since $k! \le n!$ for $1 \le k \le n$, every exponent $n!-k!$ is a non-negative integer, so $s_n \in \mathbb{Z}$. Then \begin{align*} L_n = \sum_{k=1}^{n}\frac{1}{10^{k!}} = \frac{1}{10^{n!}}\sum_{k=1}^{n}10^{n!-k!} = \frac{s_n}{10^{n!}}. \end{align*} Now evaluate the polynomial $p$ at this rational number: \begin{align*} p(L_n) &= \sum_{j=0}^{d}a_j\left(\frac{s_n}{10^{n!}}\right)^j. \end{align*} To combine these rational numbers, use the common denominator $10^{dn!}$. This gives \begin{align*} p(L_n) &= \frac{1}{10^{dn!}}\sum_{j=0}^{d}a_js_n^j10^{(d-j)n!}. \end{align*} Define \begin{align*} t_n := \sum_{j=0}^{d}a_js_n^j10^{(d-j)n!}. \end{align*} Each summand is an integer because $a_j \in \mathbb{Z}$, $s_n \in \mathbb{Z}$, and $10^{(d-j)n!} \in \mathbb{Z}$. Hence $t_n \in \mathbb{Z}$ and \begin{align*} p(L_n)=\frac{t_n}{10^{dn!}}. \end{align*} It remains to ensure that $t_n$ is not zero for all sufficiently large $n$. By the Factor Theorem, a non-zero polynomial of degree $d$ has at most $d$ roots. The sequence $(L_n)_{n=1}^{\infty}$ is strictly increasing, since \begin{align*} L_{n+1}-L_n = \frac{1}{10^{(n+1)!}}>0. \end{align*} Therefore the values $L_n$ are all distinct, and at most $d$ of them can be roots of $p$. Consequently there exists $N_0 \in \mathbb{N}$ such that $p(L_n)\ne 0$ for every $n \ge N_0$. For such $n$, the equality $p(L_n)=t_n/10^{dn!}$ implies $t_n \ne 0$. Since $t_n$ is a non-zero integer, $|t_n|\ge 1$. Therefore \begin{align*} |p(L_n)| = \frac{|t_n|}{10^{dn!}} \ge \frac{1}{10^{dn!}}. \end{align*} [/guided] [/step] [step:Compare the upper and lower bounds to force a contradiction] Choose $n \ge N_0$ with $n+1>d$. Since $p(L)=0$, the Lipschitz estimate on $[0,1]$, the tail bound, and the denominator lower bound give \begin{align*} \frac{1}{10^{dn!}} \le |p(L_n)| = |p(L_n)-p(L)| \le K|L_n-L| \le \frac{2K}{10^{(n+1)!}}. \end{align*} Multiplying by $10^{(n+1)!}$ yields \begin{align*} 10^{(n+1)!-dn!} \le 2K. \end{align*} But \begin{align*} (n+1)!-dn! = n!(n+1-d), \end{align*} and this tends to infinity along integers $n$ with $n+1>d$. Hence $10^{(n+1)!-dn!}$ is eventually larger than the fixed real number $2K$, contradicting the preceding inequality. The contradiction shows that no non-zero polynomial in $\mathbb{Z}[x]$ has $L$ as a root. Therefore the Liouville number $L$ is transcendental. [guided] We now combine the two incompatible estimates for $|p(L_n)|$. Choose $n \ge N_0$ with $n+1>d$. The lower bound from the rational denominator estimate gives \begin{align*} |p(L_n)| \ge \frac{1}{10^{dn!}}. \end{align*} Because $p(L)=0$, we can rewrite the same quantity as \begin{align*} |p(L_n)| = |p(L_n)-p(L)|. \end{align*} The numbers $L_n$ and $L$ both lie in $[0,1]$, so the Lipschitz estimate for $p$ on $[0,1]$ applies: \begin{align*} |p(L_n)-p(L)| \le K|L_n-L|. \end{align*} The tail estimate for the Liouville series gives \begin{align*} |L_n-L| \le \frac{2}{10^{(n+1)!}}. \end{align*} Combining these inequalities, \begin{align*} \frac{1}{10^{dn!}} \le |p(L_n)| = |p(L_n)-p(L)| \le K|L_n-L| \le \frac{2K}{10^{(n+1)!}}. \end{align*} Multiplying both sides by the positive number $10^{(n+1)!}$ gives \begin{align*} 10^{(n+1)!-dn!} \le 2K. \end{align*} The exponent can be rewritten as \begin{align*} (n+1)!-dn! = n!(n+1-d). \end{align*} Since $d$ is fixed and $n+1>d$, the quantity $n!(n+1-d)$ grows without bound as $n \to \infty$. Therefore the left-hand side $10^{(n+1)!-dn!}$ also grows without bound, while $2K$ is a fixed real number. This contradicts the inequality \begin{align*} 10^{(n+1)!-dn!} \le 2K \end{align*} for all sufficiently large $n$. The contradiction came only from assuming that $L$ was algebraic. Hence $L$ is not a root of any non-zero polynomial with integer coefficients, so $L$ is transcendental. [/guided] [/step]