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For the power series ${\sum }_{n\ge 0}{a}_n(z-c{)}^n$, the Cauchy-Hadamard theorem giveswith $R=\infty$ when the limit superior is zero, and $R=0$ when it is infinite. Away from the centre, the root test explains the formula: the limit superior of the term roots at displacement $z-c$ is $|z-c|/R$, interpreted as zero for $R=\infty$ and infinite for $R=0$. Values below one give absolute convergence; values above one imply that the terms fail to tend to zero. A value of one leaves the boundary question open. At the centre the series converges regardless of $R$.

For $a_{2n}=1$ and $a_{2n+1}=0$, the coefficient roots alternate between one and zero, so their ordinary limit does not exist. Their limit superior is nevertheless one, giving $R=1$ for ${\sum }_{n\ge 0}{z}^{2n}=1/(1-z^2)$ inside the unit disk. A ratio of consecutive coefficients would be undefined at the zero coefficients; the root-limsup formula handles the gap directly.

In the first part of his 1892 study of functions given by Taylor series, Jacques Hadamard introduced an upper-limit formulation for sequences and used coefficient roots to determine the radius of convergence. Sections 4–6 separate finite positive, zero, and infinite root-growth cases; the later parts study singularities and continuation. See Hadamard, Essai sur l'étude des fonctions données par leur développement de Taylor, Journal de mathématiques pures et appliquées, fourth series 8 (1892), pp. 107–109. This original text explains the coefficient-growth question without implying that the study of power series began there.

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.

The original PDF has coefficient $c_n=n^{n^{1/3}}$. Its $n$th root isThe Cauchy-Hadamard theorem therefore gives $\boxed{R=1}$. Again the power series converges absolutely inside the unit circle, while on that circle its terms do not tend to zero. The converted TeX loses the outer exponent and writes $n^{1/3}$; the calculation here uses the printed coefficient.

For the power series coefficient $c_n=n^2$, the Cauchy-Hadamard theorem givesThus the radius of convergence is $\boxed{R=1}$. The power series converges absolutely for $|z|<1$ and diverges for $|z|>1$. On $|z|=1$ its terms have modulus $n^2$, so they do not tend to zero and the series diverges there as well.

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.

At the nonzero indices $n=m!$ with $m\ge 2$,since $m/(m!)=1/(m-1)!\to 0$. At all other sufficiently large indices the coefficient is zero. The initial coefficients do not affect the limit superior, which equals one. The Cauchy-Hadamard theorem therefore gives $\boxed{R=1}$.

By Stirling's approximation,Here $(2\pi n{)}^{1/(2n)}\to 1$ and $n^{1/n^2}\to 1$, as is seen by taking the natural logarithm. The Cauchy-Hadamard theorem gives $\boxed{R=e^{-1}}$.

The Cauchy-Hadamard theorem states $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. At the prime numbers $n=p$, this root equals $p^{2/p}\to 1$. There are infinitely many prime numbers, so these indices form an unbounded subsequence. For every other positive index the root is zero. The limit superior is therefore one, and $\boxed{R=1}$.

Write the formal Maclaurin series as ${\sum }_{n\ge 0}{a}_n{x}^n$, where $a_n=f^{(n)}(0)/n!$. If the successive nonzero coefficient ratios have a limit, the ratio test givesso the series converges absolutely for $|x|<R$, whereIf that ratio does not exist, the Cauchy-Hadamard theorem gives the general formula $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. The radius of a Maclaurin series from successive derivatives concerns convergence of the series; equality to the original function additionally requires analyticity or a vanishing Taylor remainder. Mere differentiability, or even smoothness, does not ensure that equality.

For the specified logarithm,The absolute successive-term ratio tends to $|x|/2$, so the interval of absolute convergence is $(-2,2)$. At $x=2$ the nonconstant terms are the alternating harmonic series, which converges conditionally to $\ln 2$, giving the full sum $\ln 4$. At $x=-2$ all nonconstant terms are $-1/n$, so the series diverges to $-\infty$; the original logarithm is also undefined there. Thus the full real convergence interval is $(-2,2]$.

Let $a_n=(1+p/n{)}^{n^2}$. For every fixed real $p$, the base is positive once $n$ is sufficiently large; any initial negative or zero bases affect only finitely many coefficients. For that tail,The Cauchy-Hadamard theorem therefore yieldsFor example $p=0$ recovers the geometric-series radius one. The stronger expansion $a_n=e^{pn-p^2/2+O(1/n)}$ also shows that, on $|z|=e^{-p}$, the magnitudes of the individual terms tend to $e^{-p^2/2}>0$. Thus this series diverges at every point on its convergence circle; the radius itself does not require that extra boundary analysis.

For coefficients $a_n=1/\ln n$, the Cauchy-Hadamard theorem gives $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. HereThereforeOmitting the first two powers has no effect on the radius of convergence.

Apply the Cauchy-Hadamard theorem to the power series coefficients $a_n=[n/(n+\alpha ){]}^{n^2}$ for $n\ge 1$:For example, taking logarithms gives $-n\log (1+\alpha /n)\to -\alpha$. ThereforeAny convention for the isolated $n=0$ coefficient has no effect on the radius of convergence.

Write the Maclaurin series coefficients as $a_n=f^{(n)}(0)/n!$. When the successive ratios exist eventually and have a limit, the ratio test gives the displayed radius of convergence, including zero and infinity. The unconditional formula is $R^{-1}={\mathrm{limsup}}_{n\to \infty }|a_n{|}^{1/n}$, from the Cauchy-Hadamard theorem. Endpoints must be checked separately.

The existence and convergence of this power series do not by themselves imply that it equals $f$: equality requires analyticity or control of the Taylor remainder. The smooth function $f(x)=e^{-1/x^2}$ for $x\ne 0$, $f(0)=0$, has all derivatives zero at zero, so its Maclaurin series is identically zero despite the nonzero values away from zero.