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This lacunary series is analytic for $|z|<1$ and satisfies $f_l(z^l)=f_l(z)-z$. At roots of unity of $l$-power order, its radial tail diverges, giving a dense set of singularities on the unit circle and excluding algebraicity as a function. The Mahler method proves transcendence of its values at every algebraic $z$ with $0<|z|<1$. At $z=1/2$, reduced truncations have denominators $2^{l^m}$ and error less than twice their $-l$th power; for $l\ge 3$, the Roth theorem gives a shorter proof.

Write $f(z)={\sum }_{n\ge 0}{z}^{l^n}$. The authoritative PDF has exponent $l^n$; the geometric-series exponent in the converted TeX is a transcription error. On the unit disk this Fredholm series satisfiesFirst it is transcendental as a function. If ${\zeta }^{l^s}=1$, the terms with $n\ge s$ in $f(\rho \zeta )$ are the positive terms ${\rho }^{l^n}$ and their sum tends to infinity as $\rho \uparrow 1$; the finite preceding sum stays bounded. Thus every root of unity of $l$-power order is a singularity. These roots are dense on the unit circle. An algebraic function has only finitely many possible branch points or poles, as seen from its defining polynomial's leading coefficient and discriminant. Hence $f$ cannot be algebraic over $\mathbb{C}(z)$.

Suppose now that $\alpha$ is algebraic, $0<|\alpha |<1$, and $a=f(\alpha )$ is algebraic. For a positive integer $N$, a polynomial $P(X,Y)\in \mathbb{Z}[X,Y]$, nonzero and of degree at most $N$ in each variable, can be chosen so thatIndeed the first $T$ Taylor coefficients impose $T$ homogeneous integer equations on $(N+1{)}^2$ coefficients. There is a nonzero rational solution, which can be cleared to an integer one. Functional transcendence ensures $G$ is not identically zero.

Iterating the functional equation givesThus ${\beta }_k=G({\alpha }^{l^k})$ belongs to the fixed field $K=\mathbb{Q}(\alpha ,a)$ and is nonzero for all sufficiently large $k$, since a nonzero analytic $G$ has no sufficiently small nonzero zeros. Its analytic upper bound isChoose an integer $q\ge 1$ making $q\alpha ,qa$ algebraic integers, let $d=[K:\mathbb{Q}]$, and choose $M\ge 2$ bounding every conjugate of $\alpha$ and $a$. Then $q^{l^k}{\alpha }^{l^k}$ and $q^{l^{k-1}}f({\alpha }^{l^k})$ are algebraic integers. ThereforeFor every embedding $\sigma$ of $K$, the coefficient sum $L(P)$ bounds the polynomial and givesTaking the norm of the nonzero integral $D_k{\beta }_k$ yields a nonzero rational integer, of absolute value at least one. ConsequentlyOn the other hand the upper bound is $\log |{\beta }_k|\le -T\lambda l^k+O_P(1)$, where $\lambda =-\log |\alpha |>0$. Choose $N$ so large that $T\lambda >N{C}_0$, possible because $T=N^2+2N$. Letting $k\to \infty$ contradicts the two bounds. This proves, by the Mahler method, $\boxed{f(\alpha )\text{ is transcendental for }0<|\alpha |<1}$ at every algebraic $\alpha$.

For the shorter rational-approximation argument at $1/2$, truncate after $n=m$ and write $s_m=p_m/q_m$ with $q_m=2^{l^m}$. The numerator is odd because its last summand is one and all previous summands are even, so this fraction is reduced. The tail satisfiesThe value is irrational: if it were $P/Q$, the positive difference from a truncation would be at least $1/(Q{q}_m)$, contradicting the tail bound for large $m$. If it were algebraic irrational and $l\ge 3$, choose $0<ϵ<l-2$. Eventually the displayed upper bound is below $q_m^{-2-ϵ}$ for infinitely many reduced fractions. The Roth theorem says that an algebraic irrational admits only finitely many such approximations. This contradiction proves transcendence directly. The argument does not cover $l=2$, because its exponent is not strictly larger than two; the Mahler proof does cover that case.