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An analytic continuation extends a holomorphic function beyond its original domain without changing its existing values. Formally, if $U\subset V$ are domains and $f$ is holomorphic on $U$, a holomorphic function $F$ on $V$ with $F{|}_U=f$ is an analytic continuation. More locally, holomorphic functions on overlapping open sets continue one another when they agree on the overlap, allowing a chain of patches along a path. The identity theorem for holomorphic functions forces agreement throughout a connected overlap once they agree on a nonempty open set in it. It also makes an extension to a fixed connected domain unique. ^[ref]

The geometric series illustrates why a formula can reach farther than the power series that first represents it. For $z\ne 1$, its partial sum isWhen $|z|<1$, $z^{N+1}\to 0$, so the series converges to $f(z)=1/(1-z)$. The rational function $F(z)=1/(1-z)$ is holomorphic on $\mathbb{C}\setminus \{1\}$ and agrees there with the sum on the unit disc. It therefore continues $f$ to every point except its simple pole at $1$. In particular,Thus analytic continuation gives a value at $2$ without making the original geometric series converge there. Its radius of convergence remains $1$: on $|z|\ge 1$ the terms $z^n$ do not even tend to zero.

The pole at $1$ limits the original Taylor series, although most points of its convergence circle are regular for $F$. Recentring at $-1$ gives another genuinely convergent power series,At $z=-3/2$ this yields $\frac{1}{2}{\sum }_{n\ge 0}(-1/4{)}^n=2/5$, even though the original powers of $-3/2$ grow. This is the useful idea: overlapping local descriptions can represent one holomorphic function on a larger region.

Uniqueness on a connected overlap does not imply that every path leads to the same terminal germ of a holomorphic function. Continuing a complex logarithm around $0$ changes its value by $2\pi i$, giving monodromy; the return patch can carry a different branch of a multivalued function. If two patches have a disconnected overlap, agreement on one component imposes no agreement on the others. ^[ref] Continuation using the rational function $1/(1-z)$ is single-valued: any path avoiding $1$ gives the same terminal germ of a holomorphic function, even if it winds around the pole.

Knowing the radius of convergence does not settle convergence on its boundary. The three power seriesall have radius one, but behave differently on $|z|=1$.

For the geometric series, boundary terms all have complex modulus one, violating the term test for divergence. For the second series, absolute values give the divergent harmonic series at every boundary point. At $z=1$ the series itself is harmonic and diverges. At any other point on the unit circle, its geometric partial sums obeyThe Dirichlet test with the decreasing sequence $1/n\to 0$ therefore gives conditional convergence. In particular, $z=-1$ gives the negative of the alternating harmonic series, with sum $-\ln 2$. For the third series, the convergent p-series ${\sum }_{n\ge 1}1/n^2$ bounds the absolute values at every boundary point, giving absolute convergence by the comparison test for series.

At $z=-1$, the rational function $1/(1-z)$ is well defined and equals $1/2$, yet its geometric series diverges there. A formula for a function and a particular series representation can have different domains. Likewise, applying termwise differentiation of a power series preserves the radius but need not preserve the boundary convergence.

If the terms of a power series at a nonzero point $z$ are bounded by $M$, then $|a_n{w}^n|\le M(|w|/|z|{)}^n$. A convergent geometric series and the comparison test for series give absolute convergence at every point strictly inside that circle. Consequently terms at every point outside the radius of convergence must be unbounded. This conclusion is stronger than merely saying that those terms fail to tend to zero.

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.

For power series $f(z)={\sum }_{m\ge 0}{f}_m{z}^m$ and $g(x)={\sum }_{j\ge 1}{g}_j{x}^j$, their function composition is well defined as a formal power series: the coefficient of $x^n$ receives contributions only from $m\le n$, since $g$ has zero constant term. For $m\ge 1$, expand the coefficient asThe $m=0$ term is one for $n=0$ and zero otherwise. The first four coefficients of the function composition are therefore $f_0$, $f_1{g}_1$, $f_1{g}_2+f_2{g}_1^2$, and $f_1{g}_3+2f_2{g}_1{g}_2+f_3{g}_1^3$. For convergent power series, the same calculation holds wherever $g(x)$ lies inside the radius of convergence of $f$.

For a power series centred at $c$ with radius of convergence $R$, its disk of convergence is the open disk $|z-c|<R$, on which the series has absolute convergence. Boundary points may or may not belong to the larger domain of convergence. When $R=0$ this open disk is empty, although the power series still converges at its centre; when $R=\infty$ it is the whole complex plane.

The domain of convergence of a power series is the set of inputs where the series has a finite sum. Inside its radius of convergence the series has absolute convergence, and outside it diverges. At a finite positive radius the real domain may include either, both, or neither endpoint; the radius of convergence alone does not decide endpoint convergence.

In a bipartite graph, a self-avoiding walk ends in its starting class exactly when it has even length. If degrees are bounded by $\Delta$, deleting the last edge gives ${\sigma }_{2k+1}\le \Delta {\sigma }_{2k}$ and hence $Z_G^0(x)\le Z_G(x)\le (1+\Delta x)Z_G^0(x)$ for $x\ge 0$. Thus these generating functions have the same radius of convergence.

Putting $Y=e^{z}V$ in the displayed Kummer differential equation gives $z{V}^{\prime \prime }+(z+b)V^{\prime }=0$. Thus $V^{\prime }=C{z}^{-b}{e}^{-z}$ on a chosen branch, yielding $Y=e^z[A+B\int z^{-b}{e}^{-z}dz]$. For $b=1/2$ and $z=x/2$, one solution is $e^{x/2}$ and a normalized second Frobenius solution is $\sqrt{x}{\sum }_{n\ge 0}{x}^n/(2n+1)!!$. The analytic factors in both solutions have infinite radius of convergence, since their coefficient ratios tend to zero. The second solution still has a square-root branch at zero; an entire series factor does not remove a nonintegral Frobenius exponent.

This lacunary power series has radius of convergence one. For $|z|<1$, the inequality $n!\ge n$ for $n\ge 1$ gives absolute convergence by a geometric series comparison. For $|z|>1$ its terms do not tend to zero. At every point of the unit circle the terms have modulus one, so the term test for divergence proves divergence there too. The two initial terms both equal $z$, since $0!=1!=1$; this finite repetition does not affect any convergence conclusion.

A generating function packages a sequence into a series so that operations on objects become operations on their encodings. An ordinary generating function uses the power series $A(z)={\sum }_{n\ge 0}{a}_n{z}^n$; in counting problems, the coefficient $a_n$ records the number of objects of size $n$. The variable $z$ initially marks size: it need not be assigned a numerical value. Coefficient extraction writes this recovery as $[z^n]A(z)=a_n$. One compact expression can therefore answer infinitely many counting questions, while explaining how their answers fit together.

Start with two contributions: choose $i$ from $\{0,1,2\}$ and $j$ from $\{0,1\}$, with exactly one option at each allowed value. Their individual ordinary generating functions are the polynomials $P(x)=1+x+x^2$ and $Q(x)=1+x$. To count ordered pairs by total $i+j$, multiply:The coefficients $1,2,2,1$ count the ways to obtain totals zero through three. For total two, the possibilities are $(1,1)$ and $(2,0)$; the two corresponding monomials both contribute $x^2$. There are six ordered pairs altogether, agreeing with $P(1)Q(1)=3\cdot 2=6$.

If the first contribution has two distinguishable choices when $i=1$, replace $P(x)$ by $1+2x+x^2$. Its product with $Q(x)$ becomes $1+3x+3x^2+x^3$. The coefficients now carry multiplicities as well as possible totals; the exponent still records cost. This lets the same construction handle several actions with identical costs without losing their separate identities.

In general, the Cauchy product givesFor each split of the total size, the rule of product counts the choices of the two components; summing over those splits counts all possibilities. This requires each combined object to have a unique decomposition into the two components. The same idea works for costs, lengths, or other additive measures, not just the number of pieces. See MIT 18.310, generating functions, sections 1–2.

Now cover a board of length $n$ with tiles of lengths one and two, without gaps or overlaps. Let $t_n$ count these tilings. There is exactly one empty tiling, so $t_0=1$, and $t_1=1$. Removing the first tile leaves a board shorter by one or two, giving the recurrence relationThese are the Fibonacci numbers shifted by one index: $t_n=F_{n+1}$ under the convention $F_0=0$, $F_1=1$. Keeping the empty tiling is essential to both the counting rule and the resulting ordinary generating function.

For $F(z)={\sum }_{n\ge 0}{t}_n{z}^n$, the empty board or a first tile followed by a remaining tiling givesEquivalently, $F(z)={\sum }_{m\ge 0}(z+z^2{)}^m$: each factor chooses one tile, and $m$ counts pieces while the exponent of $z$ records total length. The same ordinary generating function counts ordered histories of two repeatable actions costing one and two units. Histories $(1,2)$ and $(2,1)$ differ despite having the same total cost. MIT's notes use dots and dashes with these costs, including the empty message. ^[ref]

These identities hold in formal power series algebra: each requested coefficient uses only finitely many contributions. The repeated-choice expansion is legitimate because $z+z^2$ has zero constant term; a zero-cost repeatable action would allow infinitely many histories of the same size. Numerical evaluation separately requires a convergent series. Here the radius of convergence is $(\sqrt{5}-1)/2$, the reciprocal of the golden ratio. At $z=1$, the rational function equals $-1$, but the counting series diverges: that value cannot represent its sum.

Other generating functions change the encoding: an exponential generating function divides by factorials, while a Dirichlet series uses $n^{-s}$. A counting generating function also differs from a probability generating function, whose coefficients must be nonnegative and sum to one.

For the degree-$N$ partial sum $P_N(z)=1+z+\cdots +z^N$ of the geometric series, the finite identity $(1-z)P_N(z)=1-z^{N+1}$ givesFor any fixed $0<r<1$, the triangle inequality then gives the uniform error boundEquality is attained at the positive real point $z=r$, so this is the exact worst-case error on that closed disk. It also proves uniform convergence on each such disk, not just pointwise convergence.

Given a positive tolerance $\epsilon$, choosing $N$ with $r^{N+1}/(1-r)\le \epsilon$ turns a convergence statement into a finite stopping rule. For $\epsilon =10^{-3}$, degree ten suffices throughout $|z|\le 1/2$. Nearer the boundary, $r=0.9$ requires degree $87$ and $r=0.99$ requires degree $1145$. The required degree depends on both the tolerance and the region, not just on the name of the function.

Every $R\in [0,1]$ occurs as the radius of convergence of a power series all of whose coefficients are positive integers. For $0<R\le 1$, use the ceiling function to choose $a_n=\lceil R^{-n}\rceil$. ThenFor $|z|<R$, the comparison test for series with a geometric series proves absolute convergence. For $|z|\ge R$, the bound $|a_n{z}^n|\ge (|z|/R{)}^n\ge 1$ violates the fact that terms of a convergent series tend to zero. Hence the radius of convergence is exactly $R$.

For $R=0$, use $a_n=n!$: at every $z\ne 0$ the ratio of successive term absolute values is $(n+1)|z|$, eventually exceeding two. The factorial coefficients therefore give divergence away from zero.

A power series with every coefficient a positive integer cannot have radius of convergence greater than one: $a_n\ge 1$ makes convergence impossible at $z=1$. If zero coefficients are allowed and only finitely many are nonzero, a polynomial instead has infinite radius of convergence.

Suppose $A\ne 0$ and $R>0$, and $H$ is an analytic function on a disk of radius strictly larger than $R$. The Taylor series at zero of $f(z)=A\log (1-z/R)+H(z)$ has radius of convergence $R$. Its coefficients are the sum of $-A/(n{R}^n)$ and the more rapidly decreasing coefficients of $H$. At $x=-R$, the logarithmic part is an alternating harmonic series and converges conditionally; at $x=R$ it diverges. The power series for $H$ converges absolutely at both endpoints and therefore cannot change either conclusion. The real convergence interval is $[-R,R)$.

For $f(z)=\log (2-e^z)$, take $R=\log 2$ and $H(z)=\log [(2-e^z)/(1-z/R)]$, with its branch normalized by $H(0)=0$. The quotient has nonzero removable value $2R$ at $z=R$. Its next zeros are at $R\pm 2\pi i$, so a holomorphic logarithm exists on a larger disk. This shows why the Taylor series converges at $x=-\log 2$ even though its singular positive endpoint is excluded.

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.

Integrating the geometric series for $1/(1+x)$ from zero to $x$ gives the displayed power series for $|x|<1$. It has radius of convergence one. At $x=1$ the alternating series test gives conditional convergence to $\ln 2$; at $x=-1$ the negative harmonic series diverges. Replacing $x$ by $-w$ gives $\ln (1-w)=-{\sum }_{n\ge 1}{w}^n/n$. An expression formed by substituting a nonlinear function for $w$ need not have the same convergence range after it is re-expanded as a Taylor series in the new variable.

The hyperbolic cosine is entire and $\cosh 0=1$. Continuity makes it nonzero in a neighborhood of zero, so its reciprocal is holomorphic there and has a convergent Taylor series. Since $\cosh$ and its reciprocal are even, all odd coefficients vanish.

Multiply the proposed expansion by $\cosh z$:Coefficient comparison gives $c_2+1/2=0$ and $c_4+c_2/2+1/24=0$, henceThe zeros of $\cosh z$ solve $e^{2z}=-1$, so they are $z=i(\pi /2+k\pi )$, $k\in \mathbb{Z}$. The nearest ones are $\pm i\pi /2$, and they are simple because $\sinh$ is nonzero there. The hyperbolic secant has genuine poles at these points and is holomorphic in the disk of radius $\pi /2$. The radius of convergence of its Taylor series is therefore

Start with the Maclaurin series $\cos x=1-x^2/2+x^4/24+O(x^6)$. Put $z=-x^2/2+x^4/24+O(x^6)$ into the binomial series $(1+z{)}^{-2}=1-2z+3z^2+O(z^3)$. Keeping terms through $x^4$ givessoSince $d(\tan x)/dx={\sec }^{2}x$ and $\tan 0=0$, term-by-term integration supplies the constant of integration and givesThe expansions hold near zero; the nearest real poles at $\pm \pi /2$ limit their radius of convergence to $\pi /2$. The derivative relation also independently checks all three displayed coefficients of the tangent function.

Subtract the Maclaurin series of $\log (1-x)$ from that of $\log (1+x)$. Their even powers cancel, givingThe full power series is $2{\sum }_{n\ge 0}{x}^{2n+1}/(2n+1)$, with radius of convergence one. For the real logarithm, the stated series applies on $|x|<1$.

All three examples below have radius of convergence one, because the $n$th roots of their coefficient moduli tend to one.

For convergence everywhere on the boundary, take ${\sum }_{n=1}^{\infty }{z}^n/n^2$. On $|z|=1$ it converges absolutely: for $n\ge 2$, $1/n^2\le 1/[n(n-1)]=1/(n-1)-1/n$, whose sum telescopes to a finite value.

For divergence everywhere on the boundary, take ${\sum }_{n=0}^{\infty }{z}^n$. If $|z|=1$, every term has modulus one, so the term test for divergence applies.

For mixed boundary behavior, take ${\sum }_{n=1}^{\infty }{z}^n/n$. At $z=1$ it is the divergent harmonic series; each successive dyadic block contributes at least $1/2$. At every other point $|z|=1$, summation by parts proves convergence. Indeed put $B_k={\sum }_{j=m}^k{z}^j$, so the finite geometric series gives $|B_k|\le 2/|1-z|=:C_z$. ThenThus the partial sums form a Cauchy sequence and converge. This derives the relevant Dirichlet test rather than only naming it.