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For the Airy ordinary differential equation $y^{\prime \prime }=xy$, a power series obeys $a_2=0$ and $a_{m+3}=a_m/[(m+3)(m+2)]$. Setting its first two coefficients to $(1,0)$ or $(0,1)$ givesEmpty products are one. The ratio test proves convergence on the whole complex plane. Termwise differentiation verifies the equation, and the initial normalization gives Wronskian one. By the Abel identity, the Wronskian remains one everywhere, so these solutions span the whole solution space.

Let $R_f,R_g$ be the two radii of convergence. First take $|x|<t<R_f$, with $t>0$. Absolute convergence at $t$ bounds the terms: $|a_n{t}^n|\le M$ for all $n$. ThusAt $x=0$ the right-hand series is finite. Otherwise it converges by the ratio test, because its limiting ratio is $|x|/t<1$. So the differentiated power series converges absolutely at every interior point of $f$, proving $R_g\ge R_f$. This proof also covers $R_f=\infty$ by choosing a finite $t$ above any given $|x|$; if $R_f=0$, that inequality is automatic.

Conversely, if $0<|x|<R_g$, absolute convergence givesThe constant term does not affect convergence, and both series converge at zero. Hence $R_f\ge R_g$, including the zero and infinite cases. Therefore $\boxed{R_f=R_g}$. This proof uses coefficient estimates, rather than assuming the termwise differentiation of a power series result whose radius assertion is being established.

For a series $\sum u_n$ with $u_n>0$, suppose $L={\lim }_{n\to \infty }{u}_{n+1}/u_n$ exists, allowing $L=+\infty$. The ratio test says the series converges if $L<1$ and diverges if $L>1$. In the first case an eventual ratio bound by some $q<1$ compares the tail to a geometric series. In the second case an eventual ratio above some $q>1$ prevents the terms tending to zero.

If $L=1$, the test is inconclusive: both $\sum 1/n$ and $\sum 1/n^2$ have limiting ratio one, while the first diverges and the second converges. If the ratio has no limit, this particular limit formulation gives no conclusion.

For any fixed real $x\ne 0$, the ratios between consecutive nonzero terms in the absolute-value series areandThe ratio test proves absolute convergence of both power series at every such $x$; convergence at zero is immediate. Both radii of convergence are therefore $\boxed{\infty }$. Applying the test to the nonzero terms avoids the undefined coefficient ratios caused by alternate zero coefficients when these are viewed as full power series in $x$.

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.