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A power series is lacunary when only a sparse sequence of powers has nonzero coefficients. For ${\sum }_j{c}_j{z}^{m_j}$ with strictly increasing integer exponents, the radius of convergence satisfies $R^{-1}={\mathrm{limsup}}_j|c_j{|}^{1/m_j}$ by the Cauchy-Hadamard theorem. The exponent $m_j$, rather than the term index $j$, controls the coefficient root.

The generic derivative bound implies that at every rational $q\in (0,1)$,The Cauchy-Hadamard theorem therefore gives radius zero to the Taylor series at every such $q$. If a Taylor series represented $f$ on a neighborhood of any point, that neighborhood would contain a rational point at which $f$ was real analytic, a contradiction.

For a nonzero $x$, the absolute successive-term ratio in the exponential series is $|x|/(n+1)\to 0$, so the first series converges for every $x$. For the factorial series the ratio is $(n+1)|x|\to \infty$, so its terms fail to tend to zero for every $x\ne 0$.

The third series is sparse: its exponent is $n^2$, as in the PDF. If $t_n=(n!{)}^2|x{|}^{n^2}$, thenFor $|x|>1$, the terms grow without bound; at $|x|=1$ their moduli are $(n!{)}^2$, so they again fail the term test. Equivalently, the Cauchy-Hadamard theorem for the sparse coefficients uses $(n!{)}^{2/n^2}\to 1$, not $(n!{)}^{2/n}\to \infty$. Thus the three radii of convergence areZeros at non-square coefficient indices do not change the relevant limsup in the last root test. This illustrates the radius of convergence of a sparse power series.

The equal rooted counts imply that the self-avoiding walk generating function is $Z_G(x)={\sum }_{n\ge 0}{\sigma }_n{x}^n$. The Cauchy-Hadamard theorem identifies the radius of a power series $\sum a_n{x}^n$ as the reciprocal of $\mathrm{limsup}|a_n{|}^{1/n}$. ConsequentlyEquivalently, the root test gives convergence when $x\mu <1$ and divergence when $x\mu >1$. At the positive boundary $x={\mu }^{-1}$, the preceding infimum formula gives ${\sigma }_n{\mu }^{-n}\ge 1$, so this particular series diverges there as well.

This is a lacunary power series: the coefficient of $z^{n!}$ is $n^n$, while the other coefficients are zero. Along its nonzero coefficients,The factorial dominates $n\log n$; for example $n!\ge n(n-1)(n-2)(n-3)$ for large $n$ already suffices for this limit. Hence the coefficient limsup in the Cauchy-Hadamard theorem is one. The radius of convergence is $\boxed{R=1}$.

Directly, for $|z|<1$ the consecutive term ratio is $(n+1{)}^{n+1}{n}^{-n}|z{|}^{nn!}$, which tends to zero; for $|z|\ge 1$ the term magnitudes are at least $n^n$ and fail to tend to zero. The factorial is the exponent of $z$, not part of its coefficient.

For the coefficient $a_n=1/n!$, the p-adic absolute value gives $|a_n{|}_p=p^{v_p(n!)}$. By Legendre formula,where $s_p(n)$ is the sum of the base-$p$ digits of $n$. Thus $v_p(n!)/n\to 1/(p-1)$. The Cauchy-Hadamard theorem now yieldsand therefore the radius of convergence of the p-adic exponential is

Let $L={\mathrm{limsup}}_{n\to \infty }|a_n{|}^{1/n}$. The Cauchy-Hadamard theorem, proved by applying the root test, gives a radius of convergencewith the usual conventions. For $|z|<R$, choose $q<1$ eventually bounding $|a_n{z}^n{|}^{1/n}$, which gives absolute convergence by comparison with a geometric series. For $|z|>R$, infinitely many terms have $n$th root greater than one, so the terms fail to tend to zero and the series diverges.

Here $R=2$ and the assumed limit gives $|a_n{|}^{1/n}\to 1/2$. ThereforeThus

For $f\notin E$ and rational $q\in (0,1)$, every $M$ admits $m\ge M$ withConsequentlyBy the Cauchy-Hadamard theorem, the Taylor series of $f$ at $q$ has radius of convergence zero.

Suppose some Taylor series represented $f$ throughout a neighborhood of a point $x\in [0,1]$. That neighborhood contains a rational $q\in (0,1)$. A function represented locally by a convergent power series is a real analytic function, and re-expanding that series about $q$ gives the Taylor series of $f$ at $q$ a positive radius of convergence. This contradicts the preceding conclusion. Thus $f$ agrees with no Taylor series on a neighborhood of any point; it is a nowhere-analytic generic smooth function.

By the Cauchy-Hadamard theorem, putThe squared coefficients have root limsup $L^2$, so their radius is $R^2$. Equal finite nonzero radii therefore require $R=R^2$, hence $R=1$. This can happen: for $a_n=n+1$, both $\sum a_n{z}^n$ and $\sum a_n^2{z}^n$ have radius one.

For the coefficients $a_n^{a_n}$, the relevant root terms areEquality can again occur. For example, takeThen $a_n\to \infty$, $a_n^{1/n}\to 1$, andThus both $\sum a_n{z}^n$ and $\sum a_n^{a_n}{z}^n$ have radius one. These examples illustrate the radius of convergence after powering coefficients.

For a series ${\sum }_n{a}_n{x}^{m_n}$ with distinct increasing integer exponents, define the ordinary coefficients to be $a_n$ at $m_n$ and zero elsewhere. The Cauchy-Hadamard theorem then givesThe exponent in the root is the actual degree $m_n$, not the index $n$. For $a_n=(n!{)}^2$ and $m_n=n^2$, $0\le 2\log (n!)/n^2\le 2\log n/n\to 0$, so $R=1$. Boundary behavior is a separate question.

On ${\mathbb{C}}_p$, the power series $\exp (x)={\sum }_{n\ge 0}{x}^n/n!$ has radius of convergenceIndeed, Legendre formula gives $v_p(n!)/n\to 1/(p-1)$, and the Cauchy-Hadamard theorem gives the stated radius.

The generating function weights each self-avoiding walk by $x$ to the power of its length, including the zero-length walk. Its radius of convergence is the reciprocal of the upper exponential growth rate of the rooted counts, by the Cauchy-Hadamard theorem.