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Yes: take the unnormalized sinc function with its removable value at zero,Its power series is ${\sum }_{m=0}^{\infty }(-1{)}^m{z}^{2m}/(2m+1)!$, with infinite radius of convergence, so it is an entire function. At $z=1/n$ it gives $n\sin (1/n)$ as required. The identity theorem also proves this entire solution is unique.

The geometric series for $(2+z{)}^{-1}$ converges for $|z|<2$, while the cancelled sine terms have an everywhere convergent Taylor series. Hence the requested Laurent series has no negative powers:For example,The extension through zero also makes this a Taylor series. The pole at $-2$ fixes its radius of convergence at two.

By hypothesis the power series converges at every positive real argument, so its radius of convergence is infinite. Termwise differentiation of a power series is therefore valid in a neighborhood of $s=1$. For $k\ge 1$,At $s=1$, the right side is the expectation of the falling factorial of $X$:The special case $k=1$ givesFor $k=0$, take the empty product as $1$; the equality becomes $G(1)=1$. These are the factorial moments.

The probability generating function of a nonnegative integer-valued random variable isUse $s^0=1$, including at $s=0$. The power series converges absolutely for every real $s\in [-1,1]$, since ${\sum }_{j}p(j)|s{|}^j\le 1$. In particular it is always defined on the usual probability-generating interval $[0,1]$. It may also be defined for $|s|>1$, depending on the radius of convergence and convergence at its endpoints; the precise real domain of finite expectation is $\{s\in \mathbb{R}:{\sum }_{j}p(j)|s{|}^j<\infty \}$.

Put $D=1-(1-p)(1-q)=p+q-pq$. By the law of total probability,For $0\le t<1$, termwise differentiation of the geometric series, valid inside its radius of convergence, givesTaking $t=(1-p)(1-q)$ therefore yieldsSince $D-p=(1-p)q$, this is the zero-based geometric distribution with parameterThis illustrates Bernoulli thinning of a geometric count. Here $0<p_{\ast }<1$, as $0<p,q<1$.

For the reverse conditional distribution, Bayes' theorem gives, for $n\ge k$,ThusFor $n<k$ the conditional probability is zero. The conditional distribution of $X-k$ is the failure-count form of a negative binomial distribution, with $k+1$ successes and success parameter $r$: the geometric-binomial posterior is shifted negative binomial.

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 $\infty$.

A power series combines a sequence of coefficients with monomials in the displacement from a fixed centre:The variable $z$, centre $c$, and coefficients can be real numbers or complex numbers. Its degree-$N$ partial sum $P_N(z)={\sum }_{n=0}^N{a}_n(z-c{)}^n$ is a polynomial, so a convergent power series can turn a complicated function into increasingly accurate finite calculations. Convergence, and the error after stopping, are essential parts of that claim.

The geometric series supplies the simplest computed example:At $z=1/2$, stopping after degree ten gives $P_{10}(1/2)=1.9990234375$, differing from the exact value $2$ by $1/1024=0.0009765625$. At $z=0.99$, the same degree gives only about $10.4662$ instead of $100$. The same formula has very different practical accuracy at different points; the geometric power-series remainder quantifies it.

The structure is useful beyond numerical evaluation. Within its radius of convergence, termwise differentiation of a power series turns a question about a derivative into a question about coefficients. A Taylor series chooses $a_n=f^{(n)}(c)/n!$ when it represents the function locally. A formal power series instead uses the coefficients algebraically, without asserting numerical convergence; the existing generating function examples use that distinction to encode counting problems.

For the Chebyshev differential equation $(1-x^2)y^{\prime \prime }-x{y}^{\prime }+k^{2}y=0$, a power series about the ordinary point zero splits into even and odd solutions with the displayed recurrence relation. Every nonterminating parity series has radius of convergence one. If $k^2=m^2$ for a nonnegative integer $m$, the parity series containing $x^m$ terminates, giving a polynomial with infinite radius of convergence; the other independent series still has radius one. Normalized solutions are $\cos (k\arcsin x)$ and $\sin (k\arcsin x)/k$, with the latter replaced by $\arcsin x$ when $k=0$.

For $y^{\prime \prime }-2x{y}^{\prime }+2\lambda y=0$, substitute a power series $y={\sum }_{n\ge 0}{a}_n{x}^n$ to obtain the displayed recurrence relation. It separates into an even solution with $a_0=1,a_1=0$ and an odd solution with $a_0=0,a_1=1$:The ratio of successive terms within each parity tends to zero at every fixed complex $x$, giving infinite radius of convergence. The two initial-value vectors are independent, so these solutions span the solution space. A nonzero polynomial solution exists precisely when $\lambda =N$ is a nonnegative integer: its highest power forces this value, and the recurrence terminates the matching parity. These terminating solutions are nonzero scalar multiples of the Hermite polynomials. Multiplication by $e^{-x^2/2}$ connects them to Hermite oscillator quantization.

Recognizing the displayed product derivative reduces a homogeneous second-order equation to $(py{)}^{\prime \prime }=0$. On any interval where $p\ne 0$, its general solution is $y=(A+Bx)/p$. For $p=1-x^3$, the equation becomes $(1-x^3)y^{\prime \prime }-6x^2{y}^{\prime }-6xy=0$. Data $y(0)=1$, $y^{\prime }(0)=0$ give $1/(1-x^3)={\sum }_{j\ge 0}{x}^{3j}$, with radius of convergence one. The poles occur at the three cube roots of unity. A solution singularity can disappear for a special numerator that cancels the corresponding factor, so an equation's singular point need not be a singularity of every solution.

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.

The radius of convergence of ${\sum }_{n\ge 0}{a}_n(z-c{)}^n$ is the unique $R\in [0,\infty ]$ separating interior absolute convergence from exterior divergence. It converges absolutely when $|z-c|<R$ and diverges when $|z-c|>R$. At the centre it always has the finite value $a_0$, since all positive powers vanish. The boundary convergence of a power series is a separate question.

For $0<R<\infty$, the interior is a disk in the complex plane and an open interval $(c-R,c+R)$ on the real line. A power series converges uniformly on every smaller closed disk $|z-c|\le r<R$. This gives stable polynomial approximation over a whole compact region, rather than a separate limiting calculation at each point. See Jiří Lebl, Basic Analysis, §11.3.2: convergence of power series.

Four examples reveal what $R$ measures:

The Cauchy-Hadamard theorem computes $R$ directly from the coefficients. The attractive feature is that one growth statistic decides convergence throughout the interior and exterior, including sparse series with many zero coefficients. Finite changes to the coefficients do not alter the radius, though they can change the sum.

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.

Write $a_n=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n$, $a_0=1$. The binomial series for $\sqrt{1-u}$ has positive numbers $b_n=a_{n-1}-a_n=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/[4^n(2n-1)]$ as the magnitudes of its nonconstant coefficients. The product $a_n={\prod }_{j=1}^n[1-1/(2j)]$ tends to zero, since $\ln a_n\le -\frac{1}{2}{\sum }_{j=1}^{n}1/j\to -\infty$. Thus $\sum b_n=1$ by telescoping. The Weierstrass M-test gives uniform convergence and absolute convergence on $[-1,1]$. Continuity extends the interior identity to both endpoints. Its radius of convergence is one; outside this interval its terms do not tend to zero.

Inside both convergence discs, the triangle inequality gives absolute convergence of the sum. Suppose $R<S$ and the sum had radius $T>R$. At any modulus between $R$ and $\min (T,S)$, subtracting the absolutely convergent second series from the absolutely convergent sum would give absolute convergence of the first series outside its radius of convergence, a contradiction. Interchanging the two series covers $S<R$.

For a function with derivatives of every order at $a$, its Taylor series is the power seriesIts partial sums are Taylor polynomials. At $a=0$ it is called a Maclaurin series. Forming the coefficients, proving that its partial sums form a convergent sequence, and proving that their limit equals $f(x)$ are separate tasks. A smooth function provides all the coefficients, but a real analytic function is additionally represented locally by the resulting power series.

For the exponential function, every derivative of $e^x$ at zero equals one, givingHere equality holds for every real $x$: the Taylor remainder is bounded by $e^{|x|}|x{|}^{N+1}/(N+1)!$, which tends to zero for each fixed $x$. At $x=1$, the degree-five Taylor polynomial is $163/60\approx 2.71666667$, and the absolute difference from $e$ is less than $e/720$. The factorial denominator eventually overwhelms every fixed power, allowing local derivative information to reconstruct this function throughout the real line.

The geometric series shows why the region of validity matters. For $f(x)=1/(1-x)$, the Taylor series at zero is ${\sum }_{k\ge 0}{x}^k$, with radius of convergence one. For $x\ne 1$, its finite Taylor polynomial satisfies the exact identityAt $x=1/2$ and $N=5$, the remainder is $1/32$. At $x=0.9$, the same degree-five polynomial has value $4.68559$ and remainder $5.31441$, although the Taylor series still converges to $10$ there. Convergence of the infinite expression does not guarantee accuracy from a particular small number of terms. For $|x|>1$, even the terms fail to tend to zero, so this power series cannot represent the function. The standard expansions and the conditions for convergent series are discussed in OpenStax, working with Taylor series.

A stronger caution comes from the flat functionIt is a smooth function with $h^{(n)}(0)=0$ for every $n$, so every Taylor polynomial at zero is zero and its Taylor series converges everywhere to zero. Nevertheless $h(1/2)=e^{-4}>0$. Off zero, repeated differentiation produces a polynomial in $1/x$ times $e^{-1/x^2}$; the exponential decay makes each required limit at zero vanish. This explains how all the local coefficients can miss a positive function arbitrarily close to the centre. Real analyticity supplies the additional condition that rules out this behavior.

Let a power series ${\sum }_{n\ge 0}{a}_n{z}^n$ have finite radius of convergence $R$. If its terms were bounded in complex modulus at a point $z$ with $|z|>R$, say $|a_n{z}^n|\le M$, choose $w$ with $R<|w|<|z|$. ThenThe comparison test for series with a convergent geometric series would give absolute convergence at $w$, contradicting its position outside the convergence disc. Hence the terms are unbounded at every such $z$. The argument also works when $R=0$ and gives a stronger obstruction than the usual term test for divergence.

Inside its radius of convergence, a power series $f(z)={\sum }_{n\ge 0}{a}_n(z-c{)}^n$ can be differentiated term by term:The differentiated series has the same radius of convergence. The conclusion concerns the interior, including uniform convergence on smaller closed disks; it does not assert that differentiated boundary series converge wherever the original did. For example, ${\sum }_{n\ge 1}{z}^n/n$ converges conditionally at $z=-1$, but its differentiated series ${\sum }_{n\ge 1}{z}^{n-1}$ does not.

Applied to the geometric series, it gives the useful identityDifferentiation turns the plain coefficients into integer weights. This is one way generating functions translate a weighted counting question into an operation on a known series. A proof controls the original and differentiated tails on smaller disks; the Weierstrass M-test provides the needed uniform bounds. See Lebl, Basic Analysis, §11.3: power-series differentiation and analyticity.

The sinc function $\sin z/z$ for $z\ne 0$, extended by value one at zero. The displayed power series has infinite radius of convergence, so the extension is entire. Its removable value follows from the first coefficient or from the sine limit. This convention differs by the argument scaling $z\mapsto \pi z$ from the normalized sinc convention.