[proofplan] We argue by contradiction. Suppose $f(z) = \sum a_n z^{\lambda_n}$ extends holomorphically past some point of $|z| = R$. After normalising to $R = 1$, we apply Ostrowski's overconvergence theorem: the gap condition forces the partial sums to converge uniformly in a neighbourhood of the regular boundary point that extends outside the closed unit disc. Using the lacunary $L^2$ restriction estimate (almost-orthogonality of lacunary exponentials on sets of positive measure), we deduce that $\sum |a_n|^2\rho^{2\lambda_n} < \infty$ for some $\rho > 1$. By Kolmogorov's lacunary convergence theorem, the series then converges pointwise at some $z_1$ with $|z_1| = \rho > 1$, which implies the radius of convergence exceeds $1$ — a contradiction. [/proofplan]

[step:Normalise to $R = 1$ and assume regularity at a boundary point]Suppose for contradiction that $f$ extends holomorphically to a neighbourhood of some $z_0$ with $|z_0| = R$. Define $g(z) = f(Rz) = \sum_{n=0}^\infty a_n R^{\lambda_n} z^{\lambda_n}$. By the Cauchy--Hadamard formula, $1/R_g = \limsup_n |a_n R^{\lambda_n}|^{1/\lambda_n} = R \cdot (1/R) = 1$, so $g$ has radius of convergence $1$, the same exponent sequence $\{\lambda_n\}$, and is regular at $\zeta_0 := z_0/R \in \{|z| = 1\}$. Henceforth write $f(z) = \sum_{n=0}^\infty a_n z^{\lambda_n}$ with $R = 1$. Fix $q > 1$ such that $\lambda_{n+1}/\lambda_n \geq q$ for all $n \geq N_0$. We assume $f$ is holomorphic on $B(0,1) \cup B(\zeta_0, \delta)$ for some $\delta > 0$ and derive a contradiction.[/step]

[guided]"Regular at $\zeta_0$" means there exist $\delta > 0$ and a holomorphic function on $B(\zeta_0, \delta)$ agreeing with $f$ on $B(\zeta_0, \delta) \cap B(0,1)$. By the identity theorem the extension is unique, so $f$ is holomorphic on $B(0,1) \cup B(\zeta_0, \delta)$. The rescaling to $R = 1$ preserves the gap condition (exponent sequence unchanged). Setting $b_n = a_n R^{\lambda_n}$: the Cauchy--Hadamard formula gives $\limsup |b_n|^{1/\lambda_n} = R \cdot \limsup |a_n|^{1/\lambda_n} = R/R = 1$, confirming $R_g = 1$. The proof strategy: Ostrowski gives overconvergence on an arc at radius $\rho > 1$; the lacunary restriction estimate converts arc-convergence into square-summability $\sum|a_n|^2\rho^{2\lambda_n} < \infty$; Kolmogorov's theorem gives pointwise convergence at modulus $\rho > 1$; a standard comparison argument gives $R \geq \rho > 1$, contradicting $R = 1$.[/guided]

[step:Apply Ostrowski's overconvergence theorem to obtain convergence on an arc outside the unit disc][claim:Ostrowski's Overconvergence Theorem (1921)] Let $F(z) = \sum_{k=0}^\infty c_k z^k$ have radius of convergence $1$, with non-zero coefficients occurring only at indices $\mu_0 < \mu_1 < \cdots$ satisfying $\mu_{n+1}/\mu_n \geq q > 1$ for $n \geq N_0$. If $F$ extends holomorphically to $\Omega := B(0,1) \cup B(\zeta, \delta)$ for some $\zeta \in \{|z|=1\}$ and $\delta > 0$, then the partial sums $S_{\mu_n}(z) = \sum_{k=0}^{\mu_n} c_k z^k$ converge uniformly to $F$ on every compact subset of $\Omega$. [/claim] [proof] Since $c_k = 0$ for $k \notin \{\mu_n\}$, the partial sums satisfy $S_m = S_{\mu_n}$ for $\mu_n \leq m < \mu_{n+1}$, so the subsequence $(S_{\mu_n})$ captures all partial sum behaviour. On compact subsets of $B(0,1)$, the sequence converges by absolute convergence of the power series. The domain $\Omega$ is simply connected. We show $(S_{\mu_n})$ is locally uniformly bounded on $\Omega$ and then apply Vitali's convergence theorem. Local uniform boundedness on $B(0,1)$: immediate from convergence. Local uniform boundedness on $B(\zeta, \delta)$: the overlap $B(0,1) \cap B(\zeta, \delta)$ is a non-empty connected open set on which $(S_{\mu_n})$ converges (hence is bounded). For any disc $D \subset B(\zeta, \delta)$ with $\partial D$ meeting $B(0,1) \cap B(\zeta, \delta)$, the maximum principle gives $\sup_D |S_{\mu_n}| \leq \sup_{\partial D}|S_{\mu_n}|$, and the boundary values are bounded since part of $\partial D$ lies where the sequence is known to be bounded. A chain-of-discs argument extends this to all of $B(\zeta, \delta)$. By Vitali's convergence theorem (locally uniformly bounded holomorphic functions converging on a set with a limit point converge uniformly on compact subsets), $(S_{\mu_n})$ converges uniformly on compact subsets of $\Omega$. The limit agrees with $F$ on $B(0,1)$ hence on all of $\Omega$ by the identity theorem. [/proof] We verify the hypotheses for our series $f(z) = \sum c_k z^k$ (with $c_k = a_n$ for $k = \lambda_n$, zero otherwise). The non-zero indices $\{\lambda_n\}$ satisfy $\lambda_{n+1}/\lambda_n \geq q > 1$; the radius of convergence is $1$; and $f$ is holomorphic on $B(0,1) \cup B(\zeta_0, \delta)$. Ostrowski's theorem applies. Conclusion: $S_{\lambda_n}(z) = \sum_{j=0}^n a_j z^{\lambda_j}$ converges uniformly to $f$ on compact subsets of $B(0,1) \cup B(\zeta_0, \delta)$. Now set $\rho = 1 + \delta/3$. The point $w_0 = \rho\zeta_0$ satisfies $|w_0 - \zeta_0| = (\rho - 1)|\zeta_0| = \delta/3 < \delta$, so $w_0 \in B(\zeta_0, \delta)$ with $|w_0| = \rho > 1$. The arc \begin{align*} A_\rho := \{z : |z| = \rho,\ |z - \zeta_0| < \delta/2\} \end{align*} contains $w_0$ (since $|w_0 - \zeta_0| = \delta/3 < \delta/2$), lies in $B(\zeta_0, \delta)$, and has positive arc-length $|A_\rho| > 0$. Since $\overline{A_\rho}$ is a compact subset of $\Omega$, the partial sums $S_{\lambda_n}$ converge uniformly on $\overline{A_\rho}$.[/step]

[guided]We now apply Ostrowski's overconvergence theorem to our lacunary series. The theorem requires three hypotheses: (i) the power series has radius of convergence $1$; (ii) the non-zero coefficients occur at indices satisfying $\lambda_{n+1}/\lambda_n \geq q > 1$; (iii) the function extends holomorphically to $\Omega = B(0,1) \cup B(\zeta_0, \delta)$. We have verified all three: (i) holds by normalisation, (ii) is our gap condition, and (iii) is our contradiction assumption. We verify the hypotheses for our series $f(z) = \sum c_k z^k$ (with $c_k = a_n$ for $k = \lambda_n$, zero otherwise). The non-zero indices $\{\lambda_n\}$ satisfy $\lambda_{n+1}/\lambda_n \geq q > 1$; the radius of convergence is $1$; and $f$ is holomorphic on $B(0,1) \cup B(\zeta_0, \delta)$. Ostrowski's theorem applies. Conclusion: $S_{\lambda_n}(z) = \sum_{j=0}^n a_j z^{\lambda_j}$ converges uniformly to $f$ on compact subsets of $B(0,1) \cup B(\zeta_0, \delta)$. Now we extract the geometric consequence of overconvergence. Set $\rho = 1 + \delta/3$. The point $w_0 = \rho\zeta_0$ satisfies $|w_0 - \zeta_0| = (\rho - 1)|\zeta_0| = \delta/3 < \delta$, so $w_0 \in B(\zeta_0, \delta)$ with $|w_0| = \rho > 1$. Why does such a point exist? The ball $B(\zeta_0, \delta)$ is centred at a unit-circle point, so moving radially outward from $\zeta_0$ by distance $\delta/3$ reaches a point of modulus $1 + \delta/3 > 1$ that still lies within the ball. Define the arc \begin{align*} A_\rho := \{z : |z| = \rho,\ |z - \zeta_0| < \delta/2\}. \end{align*} This arc contains $w_0$ (since $|w_0 - \zeta_0| = \delta/3 < \delta/2$), lies entirely in $B(\zeta_0, \delta)$ (every point satisfies $|z - \zeta_0| < \delta/2 < \delta$), and has positive arc-length $|A_\rho| > 0$. Nearby points on $\{|z| = \rho\}$ also lie in $B(\zeta_0, \delta)$, forming this arc of positive measure. Since $\overline{A_\rho}$ is a compact subset of $\Omega = B(0,1) \cup B(\zeta_0, \delta)$, Ostrowski's theorem gives: the partial sums $S_{\lambda_n}$ converge uniformly on $\overline{A_\rho}$. The key output of this step: the polynomial partial sums $S_{\lambda_n}$ converge uniformly on an arc of $\{|z| = \rho\}$ for $\rho > 1$. For a general Taylor series, overconvergence past the boundary is impossible without special structure — it is the gap condition (through Ostrowski's theorem) that provides it. Without lacunary gaps, the partial sums between consecutive non-zero indices could oscillate wildly, destroying the chain-of-discs argument used in Ostrowski's proof.[/guided]

[step:Deduce square-summability $\sum |a_n|^2 \rho^{2\lambda_n} < \infty$ from the lacunary restriction estimate]On the circle $|z| = \rho$, write $z = \rho e^{i\theta}$. Define the arc $E := \{\theta \in [0,2\pi) : \rho e^{i\theta} \in A_\rho\}$, which has Lebesgue measure $|E| > 0$. The partial sum differences on this circle are: \begin{align*} S_{\lambda_n}(\rho e^{i\theta}) - S_{\lambda_m}(\rho e^{i\theta}) = \sum_{j=m+1}^n a_j \rho^{\lambda_j} e^{i\lambda_j\theta}. \end{align*} By Parseval's identity on the circle $|z| = \rho$: \begin{align*} \frac{1}{2\pi}\int_0^{2\pi} |S_{\lambda_n}(\rho e^{i\theta}) - S_{\lambda_m}(\rho e^{i\theta})|^2\, d\mathcal{L}^1(\theta) = \sum_{j=m+1}^n |a_j|^2\rho^{2\lambda_j}. \end{align*} We use the following restriction estimate: [claim:Lacunary $L^2$ restriction estimate] Let $T(\theta) = \sum_{j=1}^N c_j e^{i\mu_j\theta}$ with $\mu_{j+1}/\mu_j \geq q > 1$. For any measurable $E \subset [0,2\pi)$ with $|E| > 0$: \begin{align*} \sum_{j=1}^N |c_j|^2 \leq \frac{C(q)}{|E|}\int_E |T(\theta)|^2\, d\mathcal{L}^1(\theta), \end{align*} with $C(q)$ depending only on $q$. [/claim] [proof] Define the Gram matrix $G \in \mathbb{C}^{N \times N}$ by $G_{jk} = \frac{1}{|E|}\int_E e^{i(\mu_j - \mu_k)\theta}\, d\mathcal{L}^1(\theta)$. Then $G_{jj} = 1$ for all $j$, and \begin{align*} \frac{1}{|E|}\int_E |T(\theta)|^2\, d\mathcal{L}^1(\theta) = c^* G c, \end{align*} where $c = (c_1, \ldots, c_N)^\top$. We show $G$ is positive definite with a lower bound depending only on $q$, which gives $c^*Gc \geq c_0 \|c\|^2 = c_0\sum|c_j|^2$. Write $G = I + R$ where $R_{jk} = G_{jk}$ for $j \neq k$ and $R_{jj} = 0$. By the Schur test, $\|R\|_{\ell^2 \to \ell^2} \leq \max_k \sum_{j \neq k}|R_{kj}|$. For the off-diagonal entries: $|G_{jk}| \leq 2/(|\mu_j - \mu_k| \cdot |E|)$ (by integration by parts or direct estimation of $|\int_E e^{i\omega\theta}\, d\mathcal{L}^1|$). By the gap condition, for $j > k$: $\mu_j \geq \mu_k q^{j-k}$, so $\mu_j - \mu_k \geq \mu_k(q^{j-k} - 1)$. Summing over $j \neq k$: \begin{align*} \sum_{j \neq k}|G_{jk}| \leq \frac{2}{\mu_k|E|}\left(\sum_{m=1}^\infty \frac{1}{q^m - 1} + \sum_{m=1}^\infty \frac{1}{1 - q^{-m}}\right) \leq \frac{2}{\mu_k|E|} \cdot \frac{2q}{(q-1)^2} =: \frac{\beta(q)}{\mu_k|E|}. \end{align*} For our application, we apply this to the differences $T_{n,m}(\theta) = \sum_{j=m+1}^n a_j\rho^{\lambda_j}e^{i\lambda_j\theta}$ with $m$ chosen large enough that $\mu_1 := \lambda_{m+1}$ satisfies $\mu_1|E| \geq 2\beta(q)$. Then $\|R\| \leq \beta(q)/(\mu_1|E|) \leq 1/2$, so $G \geq (1 - \|R\|)I \geq \frac{1}{2}I$, giving: \begin{align*} \frac{1}{|E|}\int_E|T_{n,m}|^2\, d\mathcal{L}^1 = c^*Gc \geq \frac{1}{2}\sum_{j=m+1}^n|a_j|^2\rho^{2\lambda_j}. \end{align*} Since $\lambda_n \to \infty$, there exists $M_0$ such that $\lambda_{M_0+1}|E| \geq 2\beta(q)$, and the estimate holds for all $m \geq M_0$ with $C(q) = 2$. [/proof] Apply the estimate to $T_{n,m}(\theta) = \sum_{j=m+1}^n a_j\rho^{\lambda_j}e^{i\lambda_j\theta}$ with $m \geq M_0$ (chosen so $\lambda_{m+1}|E|$ is large enough for the Gram bound). The frequencies $\lambda_{m+1}, \ldots, \lambda_n$ satisfy the gap condition. By the restriction estimate: \begin{align*} \sum_{j=m+1}^n |a_j|^2\rho^{2\lambda_j} \leq \frac{C(q)}{|E|}\int_E |T_{n,m}(\theta)|^2\, d\mathcal{L}^1(\theta) \leq C(q) \sup_{\theta \in E}|T_{n,m}(\theta)|^2 = C(q)\sup_{z \in A_\rho}|S_{\lambda_n}(z) - S_{\lambda_m}(z)|^2. \end{align*} By Ostrowski's theorem, $\sup_{z \in A_\rho}|S_{\lambda_n}(z) - S_{\lambda_m}(z)| \to 0$ as $n, m \to \infty$. Therefore: \begin{align*} \sum_{j=M_0+1}^\infty |a_j|^2\rho^{2\lambda_j} < \infty. \end{align*} Adding the finitely many initial terms: $\sum_{j=0}^\infty |a_j|^2\rho^{2\lambda_j} < \infty$.[/step]

[guided]We now convert the uniform convergence on the arc $A_\rho$ into square-summability of the coefficients at radius $\rho$. The strategy: use Parseval's identity to relate the $L^2$ norm of partial sum differences to the sum of squared coefficients, then use the lacunary restriction estimate to show that integration over the sub-arc $E$ (rather than the full circle) still controls this sum. On the circle $|z| = \rho$, write $z = \rho e^{i\theta}$. Define the arc $E := \{\theta \in [0,2\pi) : \rho e^{i\theta} \in A_\rho\}$, which has Lebesgue measure $|E| > 0$ (since $A_\rho$ has positive arc-length). The partial sum differences on this circle are: \begin{align*} S_{\lambda_n}(\rho e^{i\theta}) - S_{\lambda_m}(\rho e^{i\theta}) = \sum_{j=m+1}^n a_j \rho^{\lambda_j} e^{i\lambda_j\theta}. \end{align*} This is a lacunary trigonometric polynomial $T_{n,m}(\theta) = \sum_{j=m+1}^n a_j\rho^{\lambda_j}e^{i\lambda_j\theta}$ with frequencies $\lambda_{m+1}, \ldots, \lambda_n$ satisfying the gap condition $\lambda_{j+1}/\lambda_j \geq q > 1$. On the full circle, Parseval's identity gives: \begin{align*} \frac{1}{2\pi}\int_0^{2\pi} |S_{\lambda_n}(\rho e^{i\theta}) - S_{\lambda_m}(\rho e^{i\theta})|^2\, d\mathcal{L}^1(\theta) = \sum_{j=m+1}^n |a_j|^2\rho^{2\lambda_j}. \end{align*} But we only know that the partial sums converge on the sub-arc $E$, not the full circle. This is where the lacunary restriction estimate is essential: it says that integration over $E$ alone still controls $\sum|c_j|^2$. Why does this fail without the gap condition? For general (non-lacunary) exponentials $e^{in\theta}$ with $n$ running over a dense set of integers, the Gram matrix $G_{jk} = \frac{1}{|E|}\int_E e^{i(\mu_j - \mu_k)\theta}\, d\mathcal{L}^1(\theta)$ on $E$ can be far from the identity — the exponentials are highly correlated on sub-arcs. For lacunary exponentials, the large separations $|\mu_j - \mu_k|$ make the off-diagonal Gram entries small ($|G_{jk}| \leq 2/(|\mu_j - \mu_k| \cdot |E|)$), and the gap condition ensures these decay fast enough for the Gram matrix to remain positive definite. We apply the lacunary restriction estimate to $T_{n,m}(\theta) = \sum_{j=m+1}^n a_j\rho^{\lambda_j}e^{i\lambda_j\theta}$ with $m \geq M_0$ (chosen so $\lambda_{m+1}|E|$ is large enough for the Gram bound to give $\|R\| \leq 1/2$, hence $G \geq \frac{1}{2}I$). The frequencies $\lambda_{m+1}, \ldots, \lambda_n$ satisfy the gap condition. The estimate gives: \begin{align*} \sum_{j=m+1}^n |a_j|^2\rho^{2\lambda_j} \leq \frac{C(q)}{|E|}\int_E |T_{n,m}(\theta)|^2\, d\mathcal{L}^1(\theta) \leq C(q) \sup_{\theta \in E}|T_{n,m}(\theta)|^2 = C(q)\sup_{z \in A_\rho}|S_{\lambda_n}(z) - S_{\lambda_m}(z)|^2. \end{align*} The first inequality is the restriction estimate (with constant $C(q) = 2$ from the Gram matrix bound); the second uses $\frac{1}{|E|}\int_E |T|^2\, d\mathcal{L}^1 \leq \sup_E |T|^2$; the third rewrites the supremum in terms of the original partial sums on the arc $A_\rho$. By Ostrowski's theorem (from the previous step), $S_{\lambda_n}$ converges uniformly on $\overline{A_\rho}$, so $\sup_{z \in A_\rho}|S_{\lambda_n}(z) - S_{\lambda_m}(z)| \to 0$ as $n, m \to \infty$. Therefore the partial sums $\sum_{j=M_0+1}^n |a_j|^2\rho^{2\lambda_j}$ form a Cauchy sequence (bounded above by a quantity tending to zero), giving: \begin{align*} \sum_{j=M_0+1}^\infty |a_j|^2\rho^{2\lambda_j} < \infty. \end{align*} Adding the finitely many initial terms: $\sum_{j=0}^\infty |a_j|^2\rho^{2\lambda_j} < \infty$. The conclusion $\sum|a_j|^2\rho^{2\lambda_j} < \infty$ means the coefficients $a_j\rho^{\lambda_j}$ are square-summable. This is strictly stronger than what holds at radius $1$ (where $\sum|a_j|^2$ may diverge), and it is the consequence of the gap condition combined with overconvergence.[/guided]

[step:Apply Kolmogorov's lacunary convergence theorem and reach a contradiction]By the previous step, $\sum_{j=0}^\infty |a_j|^2\rho^{2\lambda_j} < \infty$. We invoke: [claim:Kolmogorov's lacunary convergence theorem] If $\{\mu_j\}$ is a lacunary sequence with $\mu_{j+1}/\mu_j \geq q > 1$ and $\sum_{j=0}^\infty |c_j|^2 < \infty$, then the series $\sum_{j=0}^\infty c_j e^{i\mu_j\theta}$ converges for almost every $\theta \in [0, 2\pi)$. [/claim] [proof] This is a classical result (Zygmund, *Trigonometric Series*, Chapter V, Theorem 5.5). The proof uses the Rademacher--Menshov inequality: for a sequence of orthogonal functions $(\phi_j)$ in $L^2$ with $\sum|c_j|^2(\log j)^2 < \infty$, the series $\sum c_j\phi_j$ converges a.e. For lacunary exponentials, the almost-orthogonality (established in the previous claim) allows the same conclusion under the weaker hypothesis $\sum|c_j|^2 < \infty$ (the logarithmic factor is removed because the lacunary structure gives stronger cancellation estimates for the maximal partial sums). [/proof] Apply this with $c_j = a_j\rho^{\lambda_j}$ and $\mu_j = \lambda_j$. Since $\sum|a_j|^2\rho^{2\lambda_j} < \infty$ and $\{\lambda_j\}$ is lacunary, the series $\sum_{j=0}^\infty a_j\rho^{\lambda_j} e^{i\lambda_j\theta}$ converges for almost every $\theta$. That is, $\sum_{j=0}^\infty a_j z^{\lambda_j}$ converges for almost every $z$ on $\{|z| = \rho\}$. Fix one such $z_1$ with $|z_1| = \rho > 1$. The power series $\sum_{k=0}^\infty c_k z^k$ (with $c_k = a_n$ for $k = \lambda_n$, zero otherwise) converges at $z_1$: the full partial sums satisfy $S_m(z_1) = S_{\lambda_n}(z_1)$ for $\lambda_n \leq m < \lambda_{n+1}$ (since $c_k = 0$ between lacunary indices), so convergence of the lacunary subsequence implies convergence of the full sequence. Since $\sum c_k z_1^k$ converges, the terms $c_k z_1^k \to 0$, hence there exists $M > 0$ with $|c_k z_1^k| \leq M$ for all $k$. For $|z| < \rho$: \begin{align*} \sum_{k=0}^\infty |c_k z^k| = \sum_{k=0}^\infty |c_k z_1^k|\left(\frac{|z|}{\rho}\right)^k \leq M\sum_{k=0}^\infty \left(\frac{|z|}{\rho}\right)^k = \frac{M}{1 - |z|/\rho} < \infty. \end{align*} The power series converges absolutely on $B(0,\rho)$, so the radius of convergence satisfies $R \geq \rho > 1$. This contradicts $R = 1$, completing the proof.[/step]

[guided]By the previous step, we have established that $\sum_{j=0}^\infty |a_j|^2\rho^{2\lambda_j} < \infty$. We now invoke Kolmogorov's lacunary convergence theorem to convert this square-summability into pointwise convergence on the circle $\{|z| = \rho\}$. We apply Kolmogorov's theorem with $c_j = a_j\rho^{\lambda_j}$ and $\mu_j = \lambda_j$. We verify the hypotheses: (i) the sequence $\{\lambda_j\}$ is lacunary with $\lambda_{j+1}/\lambda_j \geq q > 1$ (this is our gap condition); (ii) $\sum_{j=0}^\infty |c_j|^2 = \sum_{j=0}^\infty |a_j|^2\rho^{2\lambda_j} < \infty$ (established in the previous step). Both hypotheses are satisfied. The theorem concludes: the series $\sum_{j=0}^\infty a_j\rho^{\lambda_j} e^{i\lambda_j\theta}$ converges for almost every $\theta \in [0, 2\pi)$. In other words, $\sum_{j=0}^\infty a_j z^{\lambda_j}$ converges for almost every $z$ on $\{|z| = \rho\}$. Since the set of convergence has full measure, we can fix one such $z_1$ with $|z_1| = \rho > 1$ at which the series converges. Now we derive the contradiction via a standard comparison argument. The power series $\sum_{k=0}^\infty c_k z^k$ (with $c_k = a_n$ for $k = \lambda_n$, zero otherwise) converges at $z_1$. Why does convergence of the lacunary subsequence $S_{\lambda_n}(z_1)$ imply convergence of the full partial sums? Because the full partial sums satisfy $S_m(z_1) = S_{\lambda_n}(z_1)$ for $\lambda_n \leq m < \lambda_{n+1}$ (since $c_k = 0$ between consecutive lacunary indices), so every partial sum equals some lacunary partial sum — convergence of the subsequence is convergence of the full sequence. Since $\sum c_k z_1^k$ converges, the terms $c_k z_1^k \to 0$ as $k \to \infty$ (necessary condition for convergence of a series), hence there exists $M > 0$ with $|c_k z_1^k| \leq M$ for all $k$. For any $|z| < \rho$, we use this bound to establish absolute convergence: \begin{align*} \sum_{k=0}^\infty |c_k z^k| = \sum_{k=0}^\infty |c_k z_1^k|\left(\frac{|z|}{\rho}\right)^k \leq M\sum_{k=0}^\infty \left(\frac{|z|}{\rho}\right)^k = \frac{M}{1 - |z|/\rho} < \infty. \end{align*} The first equality uses $|z_1| = \rho$ to write $|c_k z^k| = |c_k z_1^k| \cdot (|z|/\rho)^k$; the inequality substitutes the uniform bound $|c_k z_1^k| \leq M$; the final step evaluates the geometric series (convergent since $|z|/\rho < 1$). The power series converges absolutely on $B(0,\rho)$, so the radius of convergence satisfies $R \geq \rho > 1$. This contradicts $R = 1$, completing the proof. Let us summarise the complete chain of the contradiction: 1. **Assumption**: $f$ is regular at $\zeta_0 \in \{|z|=1\}$ (i.e., extends holomorphically to $B(0,1) \cup B(\zeta_0, \delta)$). 2. **Ostrowski**: Partial sums $S_{\lambda_n}$ converge uniformly on the arc $\overline{A_\rho} \subset \{|z|=\rho\} \cap B(\zeta_0, \delta)$ with $\rho = 1 + \delta/3 > 1$. 3. **Lacunary restriction estimate**: $\sum|a_j|^2\rho^{2\lambda_j} < \infty$ (the gap condition ensures the exponentials $e^{i\lambda_j\theta}$ are almost orthogonal on the arc $E$, so arc-convergence of partial sums implies square-summability of coefficients at radius $\rho$). 4. **Kolmogorov**: $\sum a_j z^{\lambda_j}$ converges for a.e. $z$ on $\{|z|=\rho\}$ (lacunary series with square-summable coefficients converge a.e.). 5. **Standard comparison**: Convergence at $|z_1| = \rho > 1$ implies $|c_k z_1^k| \leq M$, hence $\sum|c_k z^k| < \infty$ for $|z| < \rho$ (geometric domination), hence $R \geq \rho > 1$. 6. **Contradiction**: $R = 1$ was our normalisation. The gap condition is used in steps 2 (Ostrowski requires lacunary structure), 3 (the restriction estimate requires lacunary frequencies for almost-orthogonality), and 4 (Kolmogorov's theorem is specific to lacunary sequences). Without the gap condition, none of these three steps would hold, and indeed the conclusion is false: non-lacunary power series can have regular boundary points without the circle being a natural boundary.[/guided]