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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.

Set $c_n=4^{-n}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$, with $c_0=1$. Cancellation of factorials givesConsequently $c_n$ decreases to zero: $\log c_n\le -\frac{1}{2}{\sum }_{j=1}^{n}1/j\le -\frac{1}{2}\log (n+1)$. The inequality uses $\log (1-x)\le -x$ and comparison with ${\int }_1^{n+1}dt/t$. Conversely $c_n\ge 1/(n+1)$ by induction, since its ratio is at least $(n+1)/(n+2)$. Thus $\sum c_n$ diverges while $\sum (-1{)}^n{c}_n$ has conditional convergence by the alternating series test. These elementary bounds require no asymptotic formula for factorials.

Standards of mathematical proof changed between the seventeenth and nineteenth centuries because both the objects studied and the operations performed on them changed. A finite geometrical construction, an infinite expansion, and an abstract group raise different questions about justification. There was no universal seventeenth-century indifference to rigour followed by its sudden nineteenth-century invention. The more defensible thesis is that successful methods repeatedly exposed unstated assumptions; later work made some of those assumptions into explicit definitions, hypotheses and existence arguments.

Pierre de Fermat's adequality illustrates the distinction between an effective method and its justification. For $F(a)=a(L-a)$, comparing neighbouring expressions givesDividing by $e$ and suppressing the remaining term in $e$ produces $a=L/2$. There is an independent finite verification:Thus the result can be secure even when the general procedure raises a question: why is the increment first nonzero for division and then discarded? In the parabola construction of question 2, the same issue appears in replacing a strict geometrical comparison by adequality and extracting the subtangent. A historical explanation must describe this operation on its own terms rather than silently equate it with a modern derivative.^[ref]

Isaac Newton and Gottfried Wilhelm Leibniz supplied different languages for the new calculus. Isaac Newton's Introduction to the Quadrature of Curves by Isaac Newton, first published with the 1704 Opticks, presents quantities generated by motion and considers their rates of change. Its mature language of first and last ratios seeks exact relations as increments vanish; it should not be confused with simply asserting that a small error is harmless. Gottfried Wilhelm Leibniz's 1684 Nova methodus pro maximis et minimis by Gottfried Wilhelm Leibniz presents a calculus of differentials and operational rules. Their common problem was to explain why procedures using vanishing increments establish exact results, while their chosen objects and modes of explanation differed.^[ref][ref]

George Berkeley's The Analyst by George Berkeley (1734) exposes the logical pressure point. For the cube,The finite identity does not become a legal division at $h=0$. The modern repair is to keep $h\ne 0$ throughout the identity and prove that its right-hand side tends to $3x^2$. That limit is different from a quotient evaluated at zero. George Berkeley's criticism therefore concerns the grounds of inference, not whether the established rules often produce correct answers. His theological polemic also reminds us that standards of mathematical proof could be contested outside specialist calculation.^[ref]

The infinite introduced another burden. Joseph Fourier's 1822 use of trigonometric series to represent arbitrary functions greatly extended what calculations in mathematical physics could express. But deriving Fourier coefficients by orthogonality only shows what those coefficients must be if the proposed operations are legitimate. It does not prove convergence, equality with the initial data, or interchange of an integral and an infinite sum. The example of a constant expanded in a Fourier sine series makes a hidden qualification visible: the terms all vanish at the endpoints, even when the assigned data do not. The correct theorem must specify the domain and mode of representation.^[ref]

Peter Gustav Lejeune Dirichlet's 1829 memoir supplies a particularly precise example of changing proof obligations. It criticises a convergence argument of Augustin-Louis Cauchy's which used complex substitution into an arbitrary function and an insufficient comparison between signed series. Peter Gustav Lejeune Dirichlet supplies the instructive pairThe first series converges by the alternating series test; the second diverges because their difference is the harmonic series. Nevertheless $b_n/a_n=1+(-1{)}^n/\sqrt{n}\to 1$. Thus asymptotic equality of terms does not transfer convergence between conditionally convergent signed series. A comparison principle valid for positive terms needs additional justification when cancellation is involved. The example refutes a mode of inference rather than merely correcting a numerical answer.^[ref]

Augustin-Louis Cauchy's own foundations programme is part of this change. The 1821 Cours d'analyse by Augustin-Louis Cauchy makes limits, continuity and series convergence central to elementary reasoning. Yet its Sum theorem of Cauchy also shows why announcing rigour does not make every proof adequate. In modern terminology, continuous functions can have a discontinuous pointwise limit. An explicit series version isEach term is a polynomial and the partial sum is $x-x^{N+1}$. The convergence does not hold in uniform convergence: with $x_N=1-1/N$, the error $x_N^{N+1}$ tends to $e^{-1}$ rather than zero. The modern uniform limit theorem repairs the inference by requiring one bound on the error to work throughout the domain. This example diagnoses the missing control under a pointwise reading; it does not settle every historical interpretation of Augustin-Louis Cauchy's infinitesimal language.^[ref]

The issue is the order of dependence in quantifiers. Pointwise convergence permits the threshold $N$ to depend on both the error tolerance and the point. Uniform convergence requires a threshold depending only on the tolerance. If $f_N$ is continuous near $x_0$ and the convergence is uniform, choose $N$ so that $|f-f_N|<\epsilon /3$ nearby, then use the continuity of this fixed $f_N$ to make $|f_N(x)-f_N(x_0)|<\epsilon /3$. The triangle inequality proves $|f(x)-f(x_0)|<\epsilon$. Writing out this proof reveals what a vague statement that the tail is small leaves unproved.

Another nineteenth-century development concerns the foundations of the real numbers themselves. Richard Dedekind's Continuity and Irrational Numbers by Richard Dedekind (1872) asks for an arithmetical account of the continuity often borrowed from a drawn line. A Dedekind cut separates the rational numbers into lower and upper classes. For instance,defines a gap in the rational numbers corresponding to $\sqrt{2}$. The cut is determined without assuming that a rational number fills it. The extended number system then supports existence theorems that a geometrical picture alone had suggested.^[ref]

This matters for apparently elementary proofs. Suppose $f$ is continuous on $[a,b]$ with $f(a)<0<f(b)$. Repeated bisection constructs nested intervals whose endpoint values have opposite signs and whose lengths tend to zero. To conclude that there is a common limiting point, one needs the completeness of the real numbers. Continuity then forces its value to be zero. Over the rational numbers, $f(q)=q^2-2$ changes sign between $1$ and $2$ but has no rational zero. Thus even a proof that avoids pictures must explain the existence of its limiting object. Richard Dedekind's account makes that requirement an explicit part of the mathematical setting, rather than an intuition about a continuous line.^[ref]

There is a parallel change in algebra. Heinrich Weber's 1882 definition in question 6 describes arbitrary elements through laws of composition. To assess such a definition one proves implications between axioms. The passage's finite set hypothesis implies that cancellative multiplication has an identity element and inverse elements; without that hypothesis, the positive integers under addition supply a counterexample. This is a new kind of proof obligation: one must establish that a proposed abstract definition has the intended consequences, rather than rely on their familiarity in particular numerical or permutation examples.^[ref]

The principal change was towards explicit control of operations, hypotheses and existence. Effective symbolic algorithms persisted, and geometric reasoning remained valuable. What became increasingly demanding was the passage from success in examples to a general claim: an infinitesimal step needed justification, an infinite expansion needed a convergence theorem, a limiting point needed a complete number system, and an abstract definition needed its consequences proved. The examples also prevent a simple progress story: nineteenth-century mathematicians disagreed about admissible reasoning, and a programme devoted to rigour could itself become the object of a successful critique.

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 $1/2$.

Use the least-upper-bound property as the least upper bound axiom: every nonempty set of real numbers bounded above has a least upper bound in $\mathbb{R}$. It implies convergence of a bounded increasing sequence: if $L$ is the supremum of its terms, then for every $\epsilon >0$ some term exceeds $L-\epsilon$, and all subsequent terms lie between that term and $L$. A bounded decreasing sequence converges by applying this argument to its negative.

The alternating series test states that if $b_n\ge 0$, $b_{n+1}\le b_n$, and $b_n\to 0$, then ${\sum }_{n\ge 1}(-1{)}^{n-1}{b}_n$ converges. To prove it, let $S_N$ denote its partial sums. The even partial sums are increasing because $S_{2m+2}-S_{2m}=b_{2m+1}-b_{2m+2}\ge 0$. The odd partial sums are decreasing because $S_{2m+3}-S_{2m+1}=-b_{2m+2}+b_{2m+3}\le 0$. Moreover,Completeness of the real numbers therefore gives a limit of a sequence for each subsequence, denoted $L_e,L_o$ respectively for the even and odd subsequences. Their difference is $S_{2m+1}-S_{2m}=b_{2m+1}\to 0$, so $L_e=L_o$. Every sufficiently late partial sum belongs to one of these two subsequences, proving convergence of the full sequence of partial sums. Multiplying the terms by $-1$ leaves convergence unchanged.

At $x=1/4$ the terms are $c_n$. Their decay to zero alone is insufficient for convergence. In factbecause the difference has numerator $n\ge 0$. Since $c_1=1/2$, induction gives $c_n\ge 1/(n+1)$. The comparison harmonic series diverges, so

At $x=-1/4$ the terms are $(-1{)}^n{c}_n$. Part (ii) gives positive decreasing magnitudes tending to zero, so the alternating series test applies. To see the convergence directly, let $S_N={\sum }_{n=1}^N(-1{)}^n{c}_n$. Thenwhile $S_{2k}\ge -c_1$, obtained by grouping $S_{2k}=-c_1+(c_2-c_3)+\cdots +(c_{2k-2}-c_{2k-1})+c_{2k}$. Thus the even partial sums have a finite limit. The odd partial sums differ from them by $-c_{2k+1}\to 0$, so the whole sequence of partial sums has the same limit. Absolute convergence would require convergence of $\sum c_n$, already disproved. Hence

The key fact is that convergence of a power series at one nonzero point forces absolute convergence at every smaller modulus. Indeed, if $\sum a_n{w}^n$ converges and $w\ne 0$, its terms form a bounded sequence, say $|a_n{w}^n|\le M$. For $|z|<|w|$,by convergence of a geometric series.

Let $S$ be the set of moduli of points at which the power series converges, and put $R=\sup S$, allowing $R=\infty$. The set contains zero, so its supremum is defined in $[0,\infty ]$. Whenever $|z|<R$, the definition of supremum provides a convergence point $w$ with $|w|>|z|$. The preceding estimate gives absolute convergence at $z$. Whenever $|z|>R$, convergence at $z$ would put $|z|$ in $S$, contradicting its upper bound. This proves the asserted radius of convergence property, including $R=0$ and $R=\infty$. Nothing in the argument decides convergence on $|z|=R$.

For the requested boundary behavior, take the power seriesIts coefficients are $a_{2k}=1/k$ for $k\ge 1$, $a_{2k+1}=0$ and $a_0=0$. At either $z=1$ or $z=-1$ it is the divergent harmonic series. At either $z=i$ or $z=-i$ it is ${\sum }_{k\ge 1}(-1{)}^k/k$, which converges by the alternating series test. In fact its radius of convergence is one, since inside the unit circle the terms are bounded by a geometric series, and outside it they do not tend to zero.

Put $s=\sin x$, so $-1\le s\le 1$. The printed summand is $(3s^n+s^{2n})/n$. If $|s|<1$, its absolute value is at most $3|s{|}^n+|s{|}^{2n}$, and the two geometric series converge. Thus there is absolute convergence.

For $s=1$ the series becomes $\sum 4/n$, a divergent multiple of the harmonic series. For $s=-1$ it becomesThe first series converges by the alternating series test, while the second diverges to $+\infty$, so their partial sums tend to $+\infty$. In particular, alternation of some terms does not imply convergence here. The complete classification isThere are no conditionally convergent cases.

Let $r=|\sin x|\in [0,1]$. For $r<1$, the absolute summand $r^n/\sqrt{n}$ is at most $r^n$, so a geometric series comparison proves absolute convergence. For $r=1$, the series is $\sum (-1{)}^n/\sqrt{n}$: its positive term sizes decrease to zero, so the alternating series test proves convergence. Its absolute series is a divergent p-series with exponent $1/2$. HenceIt converges for every real $x$ and never diverges.

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

For $s=\pm 1$, the identity $s/(2+s)=-1/3+(2/3)s$ givesThe alternating series test applies to $\sum (-1{)}^n/n$ because $1/n$ decreases to zero; its partial sums consequently have a finite limit. The harmonic series $\sum 1/n$ diverges to $+\infty$. Applying the displayed identity to each finite partial sum yieldsTherefore $\boxed{\text{the series diverges to }-\infty }$. Alternating signs alone do not satisfy the alternating series test: the magnitudes jump from $1/(3n)$ at even $n$ to $1/(n+1)$ at the next odd index.

Let $s_N={\sum }_{n=1}^N(-1{)}^{n-1}{b}_n$. Pairing consecutive terms shows $s_{2m}\ge 0$ andThe even partial sums increase and the odd partial sums decrease. They satisfy $s_{2m}\le s_{2m+1}\le b_1$, whileBoth are monotone bounded sequences and hence have limits, and the displayed difference makes their limits equal. Every partial sum belongs to one subsequence, so the alternating series converges. This is a proof of the alternating series test; the monotonicity and zero-limit hypotheses play separate roles.

The alternating series test says that if $a_n\ge 0$, $a_{n+1}\le a_n$, and $a_n\to 0$, then ${\sum }_{n\ge 1}(-1{)}^{n+1}{a}_n$ converges.

To prove it, let $S_N$ be the partial sums. The even sums satisfyso $(S_{2m})$ is increasing. The odd sums satisfyso $(S_{2m+1})$ is decreasing. Also $S_{2m}\le S_{2m+1}$, so both are bounded and converge. Their difference is $a_{2m+1}\to 0$, hence their limits agree and the whole sequence $(S_N)$ converges.

Taking $a_n=1/n$ proves convergence of the alternating harmonic series. Its even partial sums lie below its limit $S$, while its odd partial sums lie above it. Sinceandwe obtain

The alternating series test says that if $a_n\ge 0$ decreases to $0$, thenconverges. Taking $a_n=n^{-1/2}$ proves convergence of the given series.

It is not absolute convergence, because the series of absolute values is the p-serieswhich diverges since $p=1/2\le 1$. Thus the original series has conditional convergence.

For an explicit divergent rearrangement of a series, take unused positive terms, which are the even-indexed terms, until the partial sum exceeds $1$, then take the first unused negative term. Next take positive terms until the sum exceeds $2$, then the next unused negative term, and continue. Both the positive and negative subseries have infinite total magnitude, so this procedure uses every term. The negative term inserted at stage $m$ tends to zero, while the preceding partial sum exceeds $m$; consequently these rearranged partial sums tend to $+\infty$. This is a divergent series with exactly the prescribed terms.

Let $u(x)=(\log x)x^{-2/3}$. Its derivative iswhich is negative for $x>e^{3/2}$. Also $u(x)\to 0$, since a positive power eventually dominates the natural logarithm. Thus the alternating series test proves convergence of $\sum (-1{)}^{n}u(n)$.

For $n\ge 3$, $u(n)\ge n^{-2/3}$, whose p-series diverges. The comparison test therefore rules out absolute convergence. The answer is conditional convergence.

For a power series $\sum a_n{z}^n$, its radius of convergence $R\in [0,\infty ]$ separates absolute convergence for $|z|<R$ from divergence for $|z|>R$. One may define $R=\sup \{r\ge 0:\sum |a_n|r^n<\infty \}$, allowing $R=\infty$.

To justify convergence inside the radius, choose $|z|<r<R$ with $\sum |a_n|r^n<\infty$. In particular $|a_n|r^n\le M$ for some $M$, soThe comparison test with a geometric series proves absolute convergence. Conversely, convergence at a point of modulus $\rho$ makes $a_n{z}^n$ bounded there and gives absolute convergence at all smaller moduli by the same estimate. Consequently a convergent point of modulus greater than $R$ would contradict the definition of $R$.

For $0<R<\infty$, the real domain of convergence is one of $(-R,R)$, $[-R,R)$, $(-R,R]$, or $[-R,R]$. At $R=1$ these are realized respectively byThe endpoint assertions follow from the harmonic series, alternating series test, and p-series. Replacing $z$ by $z/R$ realizes every finite positive radius. The degenerate domain $\{0\}$ occurs for $\sum n!z^n$, while the entire real line occurs for $\sum z^n/n!$. Their radii are zero and infinity by the ratio test.

Here $|a_{n+1}/a_n|=n/(n+1)\to 1$. Therefore the ratio test is inconclusive. The series does in fact converge by the alternating series test, and fails to converge absolutely because the harmonic series diverges; these facts use other tests.

Put $w=e^x-1=x+x^2/2+x^3/6+O(x^4)$. The local logarithmic series $\ln (1-w)=-w-w^2/2-w^3/3+O(w^4)$ givesFor the infinite Taylor series, it is necessary to distinguish the real domain of the logarithm, $x<\ln 2$, from its convergence interval about zero. Write $a=\ln 2$. The complex zeros of $2-e^z$ are $a+2\pi ik$ for integers $k$, so the nearest logarithmic singularity to zero is at $a$. The radius of convergence is therefore $a$, giving convergence for $|x|<a$ and divergence for $|x|>a$.

The two endpoints need separate checking. Factor off the nearest singularity:The quotient has a nonzero removable value $2a$ at $z=a$; its other zeros are at $a+2\pi ik$ with $k\ne 0$. Consequently it has a holomorphic logarithm on a zero-free disk larger than $|z|\le a$, which is a simply connected domain, and the Taylor series of $H$ converges absolutely at both real endpoints. At $x=-a$, the remaining series $-{\sum }_{n\ge 1}(-1{)}^n/n$ converges by the alternating series test. At $x=a$, it is $-{\sum }_{n\ge 1}1/n$ and diverges. Thus the full real convergence set isConvergence is absolute in the open interval and conditional at the left endpoint, where the sum equals $\ln (3/2)$. This endpoint distinction is an instance of isolating a logarithmic singularity to test a power-series endpoint.