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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.

Let $a_n=n!/n^n$. Its consecutive coefficient ratio isThe ratio test for the full terms therefore has limit $|z|/e$: the series has absolute convergence when $|z|<e$ and its terms eventually grow in magnitude when $|z|>e$. Thus the radius of convergence is $\boxed{R=e}$. No boundary decision is needed to determine the radius of convergence.

A complex power series with radius of convergence $R$ converges absolutely at every $z$ with $|z|<R$ and diverges at every $z$ with $|z|>R$. This makes no general assertion at $|z|=R$: boundary points must be examined separately. The series always converges at $z=0$. If $R=0$, no nonzero point lies in the convergence disk; if $R=\infty$, the series converges absolutely throughout the complex plane.

If $p$ is the zero polynomial, every coefficient vanishes, so $R=\infty$. Otherwise let $d$ be its degree and $c\ne 0$ its leading coefficient. Since $p(n)=c{n}^d(1+o(1))$,The ratio test therefore proves absolute convergence when $|z|<1$. For $|z|>1$, the terms $p(n)z^n$ do not tend to zero, so the series diverges. Thus the polynomial-coefficient power series has $R=1$ for every nonzero polynomial. In fact, it also diverges at every $|z|=1$, because $|p(n)z^n|=|p(n)|$ does not tend to zero, whether $d=0$ or $d>0$.

Every counted point has integer coordinates and lies in the square $[-n,n{]}^2$, while the origin is always counted. HenceNo precise circle-area estimate is needed; this elementary bound controls the growth of the power series coefficients.

The comparison test for series states that if $0\le u_n\le v_n$, convergence of $\sum v_n$ implies convergence of $\sum u_n$. Equivalently, divergence of the smaller nonnegative series forces divergence of the larger one.

For $|z|=r<1$,For $0<r<1$, the ratio of successive terms on the right tends to $r<1$, so the ratio test and the comparison test for series show absolute convergence. The case $r=0$ is immediate.

If $|z|\ge 1$, then $|a_n{z}^n|\ge 1$, so the terms do not tend to zero. The term test for divergence therefore rules out convergence, including every point on the boundary $|z|=1$. Thus the radius of convergence isThe series converges exactly in the open unit disc. This is an example of polynomial coefficient growth with unit power-series radius: the lattice-count bound is polynomial rather than exponential.

Convergence at $z_0$ implies that the terms $a_n{z}_0^n$ form a bounded sequence: $|a_n{z}_0^n|\le M$ for some $M$. If $|z|<|z_0|$ and $r=|z/z_0|<1$, thenComparison with the convergent geometric series $\sum M{r}^n$ proves absolute convergence. The case $z_0=0$ is vacuous.

The radius of convergence is the number $R\in [0,\infty ]$ such that a power series converges absolutely for $|z|<R$ and diverges for $|z|>R$, with boundary behavior considered separately.

TakeThen $\sum v^n/n$ is the convergent alternating harmonic series, whereas $\sum w^n/n$ is the divergent harmonic series. Thus $\sum z^n/n$ converges at a point of modulus one and diverges at another point of modulus one. It follows that

The radius of convergence of the power series ${\sum }_{n\ge 0}{a}_n{z}^n$ isIf $|z|<R$, choose $w$ with $|z|<|w|<R$. Since $a_n{w}^n\to 0$, those terms are bounded, say $|a_n{w}^n|\le M$. Thereforeand comparison with a geometric series proves absolute convergence. If $|z|>R$, convergence would imply absolute convergence at every smaller modulus by the same argument, contradicting the definition of $R$.

If $|a_n|\le |b_n|$, convergence of $\sum b_n{w}^n$ implies absolute convergence of $\sum a_n{z}^n$ whenever $|z|<|w|$. Letting $|w|$ approach the radius of the $b$-series shows that $R_a\ge R_b$.

Assume $m^2>0$ and take $\lambda \to 0^{+}$. The free Gaussian integral gives $Z_0=\sqrt{2\pi }/|m|$, and the normalized free Gaussian measure has covariance $C=m^{-2}$. Expanding only the interaction exponential gives the asymptotic expansionIn this zero-dimensional scalar field theory, the path integral has become an ordinary integral, but its combinatorics are precisely those of sextic scalar field theory.

The free generating functional is $⟨e^{J\varphi }{⟩}_0=e^{C{J}^2/2}$. Differentiating it proves Wick theorem: odd moments vanish, while each even moment is the sum over pairings of its factors. Each Wick contraction contributes $C$, soThus the Feynman rules are a quantum field theory propagator $C$ for each internal line and a factor $-\lambda$ for each six-valent interaction vertex. The factorials $V!$ and $(6!{)}^V$ remove vertex and half-edge labels; the remaining weight is $(-\lambda {)}^V{C}^I/S$, where $S$ is the Feynman-diagram symmetry factor, the order of the automorphism group of the resulting diagram, including its half-edge symmetries. A tadpole diagram has an additional interchange symmetry of the two ends of its loop.

There is one empty Vacuum Feynman diagram, contributing $1$. At one interaction vertex, the six half-edges form three loops. Its Feynman-diagram symmetry factor is $2^{3}3!=48$, so its contribution is $-\lambda C^3/48$.

At two interaction vertices let $r$ count the lines joining them. Each interaction vertex has $(6-r)/2$ self-loops, so $r$ can only be $0,2,4,6$. These four possibilities exhaust all Vacuum Feynman diagrams at this order. Their Feynman-diagram symmetry factors areHere the leading $2$ exchanges the two interaction vertices, $r!$ permutes the joining lines, and the remaining factors permute and reverse the self-loops. The $r=0$ graph is disconnected and must be included in $Z/Z_0$; only a connected generating functional would discard it.

Consequently,Indeed ${\sum }_r{S}_r^{-1}=77/7680$, independently agreeing with $11!!/[2(6!{)}^2]$ from Wick theorem.

This is an asymptotic expansion, rather than a convergent perturbation series: the coefficient of ${\lambda }^V$ contains $(6V-1)!!/[V!(6!{)}^V]$ and grows too rapidly for a nonzero radius of convergence. For any fixed truncation order, the Taylor expansion remainder bound for $e^{-x}$ on $x\ge 0$, followed by the finite Gaussian moment integral, proves the stated remainder estimate. For negative real $\lambda$ the original integral diverges.

LetThe set $S$ contains zero. If $r\in S$ and $0\le s<r$, then $s\in S$ by comparison. Moreover, if the original power series diverges at $x_0$, no $r>|x_0|$ can lie in $S$, because absolute convergence at $r$ would imply absolute convergence at $x_0$. Thus $R=\sup S$ is finite and nonnegative.

If $|x|<R$, choose $r\in S$ with $|x|<r$; comparison gives absolute convergence at $x$. If $|x|>R$ and the series converged at $x$, then its terms $a_n{x}^n$ would be bounded. For any $y$ with $R<|y|<|x|$,so the geometric series comparison would put $|y|$ in $S$, contradicting the definition of $R$. This proves the radius of convergence property.

For ${\sum }_{n\ge 1}{x}^n/3^n$, the ratio is $x/3$, so the geometric series criterion gives

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

The Taylor series of the complex sine isTherefore, for $z\ne 0$,and the right side defines a holomorphic function at $z=0$ with value $1$. Thus the apparent singularity is a removable singularity.

The resulting power series isThe ratio test shows convergence for every $z\in \mathbb{C}$, so its radius of convergence is

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.

The exact energies and mixing coefficients containAs a function of complex $\lambda$, its nearest branch points to the origin occur when its argument vanishes:Their distance from the expansion point $\lambda =0$ determines the radius of convergence. ThereforeThe real-axis perturbation series converges for $|\lambda |<(E_{+}-E_{-})/2$.

Suppose real coefficients satisfy $c\le a_n\le C(1+n{)}^m$ for all sufficiently large $n$, where $c,C>0$ and $m\ge 0$. Then the power series $\sum a_n{z}^n$ has radius of convergence one. For $|z|<1$, the comparison test for series and the ratio test dominate it by a polynomial times a geometric sequence. For $|z|\ge 1$, the terms have modulus at least $c$ eventually and violate the term test for divergence. Finite changes to the coefficients do not affect the conclusion.

For a nonzero polynomial $p$, the power series ${\sum }_{n\ge 0}p(n)z^n$ has radius of convergence $1$. If $p$ has degree $d$ and leading coefficient $c\ne 0$, then $p(n)\sim c{n}^d$ and $|p(n+1)/p(n)|\to 1$, so the ratio test applies inside the unit disk. At every point of its boundary, the terms fail to tend to zero. The zero polynomial is the exceptional case with radius of convergence $\infty$.

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.