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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.

The positive sequence $a_n=\sqrt{n}(n/e{)}^n/n!$ has logarithmic incrementsThe rational lower bound for the natural logarithm proves $u_n>0$. For $n>1$, the Taylor series givesSince $0<(k-1)/[2k(k+1)]<1/2$, a geometric series bounds $u_n<1/[2n(n-1)]\le n^{-2}$ for $n\ge 2$. The comparison test for series therefore proves $\sum u_n<\infty$. The finite telescoping sum $\ln a_{n+1}=-1+{\sum }_{k=1}^n{u}_k$ shows that $a_n$ increases to a finite positive limit. This argument establishes the existence of the constant in Stirling's formula; its value $1/\sqrt{2\pi }$ requires an independent identification.

A geometric series adds the terms of a geometric progression:Its ordinary infinite sum is the limit of a sequence of partial sums $S_N={\sum }_{j=0}^{N-1}a{r}^j$, rather than a new operation that adds infinitely many terms at once. For $a,r$ that are real numbers or complex numbers, if $|r|<1$, the series is absolutely convergent andWhen $a\ne 0$, this condition on the modulus of $r$ is also necessary: for $|r|\ge 1$ the terms fail to tend to zero. The all-zero series $a=0$ converges for every $r$. The finite geometric series identity and $r^N\to 0$ give the result with almost no extra algebra. ^[ref]

With $a=1$ and $r=1/2$, the partial sums approach $2$ from below. Changing the common ratio to $-1/2$ makes the partial sums approach $2/3$ from alternating sides: their increments still shrink. At $r=-1$, those increments never shrink, and the partial sums remain $0,1,0,1,\dots$. Substituting $r=-1$ into $1/(1-r)$ does not supply the ordinary sum of that divergent series.

The formula is useful because many apparently different problems share a fixed multiplier. A repeating decimal such as $0.\overline{27}$ isA geometric distribution counting failures before the first success normalizes by the same identity: for $0<q<1$, the probabilities $(1-q)q^n$ on $n=0,1,\dots$ add to one. The comparison test for series uses a bound by a decaying geometric progression to establish absolute convergence without an exact formula for the series being studied. The iterated contraction gives a summable orbit result uses precisely this idea for repeated applications of a map.

The remainder of a geometric series gives more than an eventual limit: it states exactly how much an approximation omits. A common ratio close to one still gives a convergent series, but can require many terms for a useful approximation.

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.

Let $M$ bound all $|a_n{w}^n|$ and put $q=|z|/|w|<1$. ThenThe geometric series ${\sum }_{n=0}^{\infty }M{q}^n$ converges, so the comparison test for series proves absolute convergence of $\sum a_n{z}^n$. This is the bounded power-series terms force interior absolute convergence principle.

Define the radius of convergence byallowing $\rho =0$ and $\rho =\infty$. If $|z|<\rho$, the supremum supplies a point of convergence $w$ with $|w|>|z|$; convergence makes its terms bounded, so the argument above gives absolute convergence at $z$. If $|z|>\rho$, convergence is impossible by definition. At $|z|=\rho$, no general conclusion follows. Thus the convergence region has a circular interior centered at zero, with boundary behavior depending on the power series.

Group three consecutive terms. Their block sum isSince $2n-1\ge n$ and $4n-1\ge 3n$ for $n\ge 1$, we have $|b_n|\le 1/(2n^2)$. The comparison test for series and the preceding proof of convergence of $\sum n^{-2}$ show that $\sum b_n$ converges absolutely.

This establishes convergence of the original partial sums at indices $3N$. Each intermediate partial sum differs from one of these by at most two of the original terms; those terms tend to zero. Thus all partial sums have the same limit of a sequence, which is the mechanism of convergence from fixed-length series blocks. The original series is not absolutely convergent, since its absolute-value series contains ${\sum }_n(2n-1{)}^{-1}\ge \frac{1}{2}{\sum }_n{n}^{-1}$. Therefore the original series converges conditionally.

The positive terms $b_n=1/\log (n+1)$ decrease to zero. The alternating series test just proved therefore gives convergence of the signed series. It is conditional convergence: $\log (n+1)\le n$, so $b_n\ge 1/n$, and the absolute-value series dominates the divergent harmonic series.

The comparison test for series used here says that if $0\le c_n\le d_n$, convergence of $\sum d_n$ implies convergence of $\sum c_n$, while divergence of $\sum c_n$ implies divergence of $\sum d_n$. This follows by comparing increasing partial sums. For completeness, the harmonic series diverges because every block from $2^k$ to $2^{k+1}-1$ has sum at least $1/2$.

The absolute ratio of successive terms is $|z|n^2/(n+1{)}^2\to |z|$. The ratio test gives convergence for $|z|<1$ and divergence for $|z|>1$, so the radius of convergence is $\boxed{R=1}$. On the boundary, $|n^{-2}{z}^n|=n^{-2}$ andThe positive partial sums are bounded, and thus converge. By the comparison test for series, the original power series converges absolutely at every point of $|z|=1$.

Suppose, towards a contradiction, that $|a_n{z}^n|\le M$ for every $n$, where $|z|>R$. Choose a radius $\rho$ strictly between $R$ and $|z|$, and take any $w$ with $|w|=\rho$. ThenThe geometric series on the right converges because its ratio lies in $[0,1)$. The comparison test for series therefore gives absolute convergence of the power series at $w$, contrary to the defining property of its radius of convergence. Thus $|a_n{z}^n|$ is unbounded in $n$ whenever $|z|>R$. This is the bounded power-series terms force interior absolute convergence principle. The case $R=\infty$ has no points outside its convergence disk and the assertion is vacuous there.

Every possible radius of convergence is realized by an explicit power series:For the first, consecutive absolute terms have ratio $(n+1)|z|$, so at every $z\ne 0$ they eventually grow and do not tend to zero; at zero the series is defined. The middle geometric series converges absolutely exactly for $|z|<R$, and outside that disk its terms grow. The last has consecutive-term ratio $|z|/(n+1)\to 0$, so the ratio test gives absolute convergence for every finite $z$. These arguments use the necessary zero-term condition for convergence, the comparison test for series, and absolute convergence of a geometric series with ratio less than one; they do not assume convergence on the boundary.

For odd $n$, the denominator is $(n+1{)}^2$; for even $n$, it is $(n-1{)}^2$. There is no zero denominator in the printed index range. For all $n\ge 2$,The comparison test for series with the convergent p-series gives absolute convergence. In fact the consecutive pair $n=2k-1,2k$ contributes $1/(2k{)}^2+1/(2k-1{)}^2$, so the sum is the same as ${\sum }_{m=1}^{\infty }{m}^{-2}$. Positivity justifies this paired reordering; an alternating-sign appearance in the denominator does not make this an alternating series.

The two-variable arithmetic-geometric mean inequality follows from a square:Hence $2\sqrt{ab}\le a+b$, with equality exactly when $a=b$.

For the series, apply that same inequality with $a=a_n$ and $b=n^{-2}$ to obtainThe comparison series converges: its first summand has finite sum by hypothesis, and the inverse-square series converges, for example by the telescoping estimate $n^{-2}\le 1/[n(n-1)]=1/(n-1)-1/n$ for $n\ge 2$. Thus the comparison test for series provesEquivalently, its positive partial sums are increasing and bounded above by $\frac{1}{2}\sum a_n+\frac{1}{2}\sum n^{-2}$, so the bounded monotone sequence theorem proves convergence directly. This is the case $p=1$ of weighted square roots of a summable sequence.

For any fixed $r>0$, $\log y=o(y^{1/r})$ as $y\to \infty$. For example, l'Hôpital's rule applied to $\log y/y^{1/r}$ gives the limit zero. With $y=\log n$, this implies $(\log \log n{)}^r<\log n$ eventually. Thereforefor all sufficiently large $n$. The comparison series diverges by the integral test for convergence, sinceThe comparison test for series givesOne may obtain the same growth bound by extending the given logarithmic estimate to real $y$ through $\lceil y\rceil$ and choosing any exponent smaller than $1/r$; no special convergence test for iterated logarithms is needed.

The comparison test for series says that an eventual upper bound by a convergent positive series proves convergence, while an eventual lower bound by a divergent positive series proves divergence. The p-series $\sum n^{-s}$ converges exactly when $s>1$.

If $p>1$, then $(\log n{)}^q\ge 1$ for all sufficiently large $n$, soand the series converges. If $0<p<1$, choose $\eta =(1-p)/2>0$. The allowed logarithmic bound, applied with exponent $\eta /q$, gives $(\log n{)}^q<n^{\eta }$ eventually. Thusand the series diverges.

For $p=1$, use the integral test for convergence: for an eventually positive decreasing function $g$, $\sum g(n)$ and $\int g(x)dx$ either both converge or both diverge. Here $g(x)=1/[x(\log x{)}^q]$ is decreasing for $x>1$, and the substitution $u=\log x$ givesThis integral converges exactly when $q>1$. Hence the polynomial-logarithmic series convergence classification isAll other allowed positive $p,q$ give divergence.

A power series has radius of convergence $R$ when it converges absolutely for every $|z|<R$ and diverges for every $|z|>R$. Its behaviour on $|z|=R$ is not specified by this definition. When $R=0$, it converges at the centre and diverges at every nonzero point; an infinite radius of convergence means absolute convergence at every finite $z$.

Suppose, towards a contradiction, that $|a_n{z}^n|\le M$ for all $n$ at some $|z|>R$. Choose a complex number $w$ with $R<|w|<|z|$. ThenThe comparison test for series with the convergent geometric series $\sum M{q}^n$ yields absolute convergence at $w$, contradicting the definition of $R$. Thus terms of a power series are unbounded outside its convergence disc:For the given power series, the ratio test gives, when $z\ne 0$,Hence it converges absolutely for $|z|<1$ and its terms fail to tend to zero for $|z|>1$ (the ratios are eventually larger than some number greater than $1$). Therefore $\boxed{R=1}$. In fact, on $|z|=1$ it also converges absolutely by comparison with the convergent p-series $\sum n^{-3}$.

The ratio test says that a series $\sum u_n$ whose terms are eventually nonzero converges absolutely if $\mathrm{limsup}|u_{n+1}/u_n|<1$; in particular, it suffices to have $|u_{n+1}/u_n|\le q<1$ eventually, since then its terms are bounded by a convergent geometric series. The comparison test for series states that, for nonnegative terms, comparison with a larger convergent series gives convergence, and comparison with a smaller divergent series gives divergence. For $u_n=n!/n^n$,By the binomial theorem, $(1+1/n{)}^n\ge 1+n(1/n)=2$. Thus $u_{n+1}/u_n\le 1/2$ for every $n\ge 1$, and induction gives $0<u_n\le 2^{1-n}$. The comparison test for series with $\sum 2^{1-n}=2$ proves

The limit comparison test says that if $u_n,v_n>0$ and $u_n/v_n\to L$ with $0<L<\infty$, then their series either both converge or both diverge. This follows because, eventually, $(L/2)v_n\le u_n\le (3L/2)v_n$, and the comparison test for series applies in both directions.

Here comparison with the harmonic series givesFor completeness, the natural logarithm grows slowly enough because for $x\ge 1$,so $(\log n{)}^2/n\le 16/\sqrt{n}\to 0$. The harmonic series diverges: every block from $n=2^k$ to $2^{k+1}-1$ has $2^k$ terms at least $2^{-k-1}$ and therefore contributes at least $1/2$. Hence

The alternating series test states that if $b_n\ge 0$ decreases to zero, then $\sum (-1{)}^n{b}_n$ converges. Indeed, its even and odd partial sums are monotone in opposite directions and bound each other, while their difference tends to zero. Here $b_n=1/(1+\sqrt{n})$ decreases to zero, so the series converges.

It fails absolute convergence, since for $n\ge 1$,The p-series $\sum n^{-1/2}$ diverges. This also follows by comparison with the divergent harmonic series, because $n^{-1/2}\ge n^{-1}$. The comparison test for series therefore shows that $\sum b_n$ diverges. Thus

First, for $0<x<1$, the derivative of $h(x)=\log (1-x)+x$ is $-x/(1-x)<0$, while $h(0)=0$. Hence $\log (1-x)<-x$.

Choose $N>c$ large enough that the given ratio bound holds for every $n\ge N$. Positivity permits taking logarithms, and the preceding inequality givesSumming, and using the integral test for convergence from part (b), gives for $n>N$Thus $a_n\le a_N{N}^c{n}^{-c}$. Since $c>1$, the P-series $\sum n^{-c}$ converges, so the comparison test for series provesThe finitely many initial terms do not affect convergence. This power-law comparison from a refined ratio bound supplies information when the ordinary ratio test can be inconclusive.

Use $\sin 0=0$, the derivative identity ${\sin }^{\prime }=\cos$, continuity of cosine at zero, and $\cos 0=1$. For $u\ne 0$, the mean value theorem gives $\sin u/u=\cos c_u$ for a point between zero and $u$. Hence $\sin u/u\to 1$ as $u\to 0$. For $x\ne 0$, apply this with $u=x/3^k$ to obtain the limit $\boxed{3^k\sin (x/3^k)\to x}$. For $x=0$ every term is zero.

The additional bound $|\cos t|\le 1$ and the same mean value theorem give $|\sin u|\le |u|$. ThereforeThe right side is summable. By the comparison test for series the stated series has absolute convergence for every real $x$.