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The alternating harmonic series ${\sum }_{n\ge 1}(-1{)}^{n+1}/n$ converges. Its even partial sums increase, its odd partial sums decrease, and the two limits agree because their difference is the next term. In particular, every even partial sum is a lower bound and every odd partial sum is an upper bound for the limit.

This alternates the signs of the terms of the harmonic series, which itself diverges.

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.

For a finite set of primes,expands as the sum of $1/n$ over positive integers whose prime factors are at most $x$. It contains every term with $n\le x$, so these partial products dominate the harmonic series and diverge. Sinceuniformly for $0\le t\le 1/2$, convergence of ${\sum }_{p}1/p$ would force convergence of the logarithms of these products, a contradiction.

The harmonic series is the positive seriesIts partial sums are the harmonic numbers $H_N$. Although the individual terms tend to zero, $H_N$ grows without bound, so the harmonic series is not a convergent series. This is a basic warning against using the size of one term as a bound on all the terms still to come.

The growth is slow enough to be visually deceptive. The partial sums at $N=10,100,1000,10000$ are approximately $2.92896825,5.18737752,7.48547086,9.78760604$. Adding ten times as many terms increases the accumulated total by roughly $\log 10\approx 2.30258509$, rather than multiplying the total by ten. The telescoping series $\sum 1/[n(n+1)]$ has similarly vanishing terms but approaches the finite sum one.

A short block argument proves the divergence without trusting a finite plot. For each integer $k\ge 0$, the block from $2^k+1$ through $2^{k+1}$ contains $2^k$ terms, each at least $1/2^{k+1}$. ThereforeEvery doubling supplies at least another half, and arbitrarily many doublings are possible. A collection of tiny terms can still have a substantial combined contribution. See Jiří Lebl, Basic Analysis I, Example 2.5.11 for this grouping proof.

Integral comparison sharpens the growth estimate. Here $\log$ denotes the natural logarithm. Since $1/x$ is a strictly decreasing function for positive $x$, comparison of rectangles with the integral givesThus the divergence is logarithmic. The finer difference $H_N-\log N$ tends to the Euler--Mascheroni constant, approximately $0.57721566$; the harmonic number expansion supplies higher-order corrections. This turns the counterexample into a usable quantitative estimate of accumulated reciprocal contributions.

Euler's De progressionibus harmonicis observationes, written in 1734 and published in 1740, studies these sums beyond the yes-or-no question of convergence. Sections 2–3 compare long blocks to show continued growth, and later sections compare the sums with logarithms and calculate the constant offset. The ^[ref] identifies the dates and original publication; ^[ref] makes the calculations accessible. This is not a claim that Euler first discovered divergence: the paper's interest includes measuring how slowly a divergent sum grows.

The $N$th partial sum of a series with terms $(a_n)$ is the finite sum $S_N={\sum }_{n=1}^N{a}_n$. It records the accumulated total after $N$ additions. The terms and their partial sums are different sequences: the terms of the harmonic series get small, while their partial sums grow without bound.

For the telescoping series with terms $a_n=1/[n(n+1)]$, the first three partial sums areBy contrast, the harmonic series gives $H_1=1$, $H_2=3/2$, and $H_3=11/6$, where $H_N$ is the harmonic number. Computing more partial sums approximates an infinite sum only when a limit of a sequence exists, and the tail of a telescoping reciprocal series shows how an error estimate can certify the approximation.

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.

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 $1/2$.

First prove the Baire category theorem in its complete-metric form. Let $X$ be a nonempty complete metric space, let $G_n$ be open dense sets, and let $V$ be any nonempty open set. Choose a closed ball $\overline{B}(x_1,r_1)\subset V\cap G_1$ with $0<r_1<1/2$. Inductively, density and openness permit the choiceFor $m>n$, $x_m$ belongs to the $n$th closed ball, so $d(x_m,x_n)\le r_n$. The centers form a Cauchy sequence and converge to some $x\in X$ by completeness. Each closed ball contains all later centers and therefore contains their limit. Hence $x\in V\cap {\bigcap }_n{G}_n$. Since this holds for every $V$, every countable intersection of open dense sets is dense. Equivalently, a nonempty complete metric space cannot be a countable union of closed sets with empty interior; this is the form needed below.

For the Banach-Steinhaus theorem, let $X$ be a Banach space, $Y$ a normed vector space, and $\mathcal{T}$ a family of bounded linear operators from $X$ to $Y$ such that ${\sup }_{T\in \mathcal{T}}\Vert Tx\Vert <\infty$ for every $x$. DefineEach $E_m$ is closed, as an intersection of inverse images of closed balls under continuous maps, and pointwise boundedness gives $X={\bigcup }_m{E}_m$. The Baire category theorem gives some $E_m$ with nonempty interior. Choose $B(x_0,r)\subset E_m$. If $\Vert h\Vert \le r/2$, both $x_0+h$ and $x_0$ lie in $E_m$, so $\Vert Th\Vert \le 2m$ for every $T$. Applying this to $h=(r/2)u$ with $\Vert u\Vert \le 1$ provesThis proves the Uniform boundedness principle; completeness is required for the domain, but not for $Y$.

Now fix the prescribed point ${\theta }_0$ on the unit circle, parametrized by angles modulo $2\pi$. On the real Banach space $C(\mathbb{T};\mathbb{R})$ with the supremum norm, consider the continuous linear functionals $T_{N}f=S_{N}f({\theta }_0)$, where $S_N$ is the symmetric Fourier partial sum. Summing the finite geometric series gives its Dirichlet kernel representationThe continuous extension at $t=0$ is $2N+1$. The kernel bound gives $\Vert T_N\Vert \le {\Lambda }_N=(2\pi {)}^{-1}\int |D_N|$. Equality holds: the continuous periodic functions defined byhave supremum norm at most one, and dominated convergence gives $T_N{f}_{\delta }\to {\Lambda }_N$ as $\delta \downarrow 0$.

These norms are unbounded. Indeed, evenness and $\sin (t/2)\le t/2$ for $0<t\le \pi$ giveFor the second inequality, integrate over $[(j-1)\pi ,j\pi ]$, where $1/u\ge 1/(j\pi )$ and the integral of $|\sin u|$ is two. The harmonic series diverges, proving the Dirichlet kernel harmonic lower bound. If every continuous $f$ had bounded partial sums at ${\theta }_0$, the Banach-Steinhaus theorem would bound all the operator norms, a contradiction. Consequently there exists a continuous real function $f$ with ${\sup }_N|S_{N}f({\theta }_0)|=\infty$, and its Fourier series diverges at the prescribed point. The function depends on the chosen point; the argument requires no claim of simultaneous divergence everywhere.

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

Let $a_n$ denote the summand. Its denominator is positive for every $n\ge 1$: odd $n$ gives $n^4+n^2$, while even $n\ge 2$ gives $n^4-n^2$. Dividing numerator and denominator by their highest powers givesThe terms are positive for sufficiently large $n$, so the limit comparison test with the divergent harmonic series gives divergence to $+\infty$. More explicitly, for $n\ge 16$ the numerator is at least $n^3/2$ and the denominator is at most $2n^4$, giving $a_n\ge 1/(4n)$. Any finitely many early negative terms change partial sums by only a finite constant and cannot prevent divergence of this positive tail.

Expand the finite Euler product into convergent geometric series. Every positive integer $n\le x$ has all its prime factors at most $x$, soThe harmonic series diverges, hence $P(x)\to 0$. If the sum of reciprocal primes were finite, then the inequality $-\log (1-t)\le 2t$ for $0\le t\le 1/2$ would giveuniformly in $x$, contradicting $P(x)\to 0$. Thus the sum of prime reciprocals diverges. Unique factorization justifies the expansion; no estimate for the distribution of primes is required. This is the Euler proof that the sum of reciprocals of primes diverges.

First establish a comparison fact for a power series. If it converges at $w\ne 0$, its terms $a_n{w}^n$ are bounded, say $|a_n{w}^n|\le C$. For $|z|<|w|$,by the geometric series. Thus convergence at one point implies absolute convergence at every point of smaller modulus.

Let $S$ be the set of moduli of points at which the power series converges. It contains zero. Define the radius of convergence $R=\sup S\in [0,\infty ]$. If $|z|<R$, the definition of supremum gives a convergent point $w$ with $|w|>|z|$; the comparison just proved gives absolute convergence at $z$. This reasoning also applies when $R=\infty$. If $|z|>R$ and the power series converged there, its modulus would belong to $S$, contradicting the definition of $R$. ThereforeThe boundary $|z|=R$ is intentionally undecided by this argument. If $R=0$, only the exterior assertion is nonvacuous.

For the particular coefficients $a_n=1/n$ for $n\ge 1$, comparison with $\sum |z{|}^n$ gives absolute convergence whenever $|z|<1$. If $|z|=r>1$, then $r^n/n$ fails to tend to zero: its successive ratio is $rn/(n+1)$, eventually greater than a fixed number bigger than one. The necessary term condition for convergence of a series therefore fails. HenceFor instance the positive boundary point gives the divergent harmonic series; no assertion of universal boundary convergence is needed.

A real sequence $x_n$ converges to $l$ if, for every $\epsilon >0$, there is an integer $N$ such that $|x_n-l|<\epsilon$ for every $n\ge N$. A convergent series ${\sum }_{n=1}^{\infty }{x}_n$ is one whose partial sums $s_N={\sum }_{n=1}^N{x}_n$ converge to a finite real limit.

If $s_N\to s$, then $x_N=s_N-s_{N-1}\to s-s=0$. This necessary condition is the term test for divergence. Its converse fails: the harmonic series has terms $1/n\to 0$ but diverges. For a direct proof, each block from $n=2^{j-1}+1$ to $2^j$ contains $2^{j-1}$ terms each at least $2^{-j}$, so contributes at least $1/2$. Infinitely many such blocks make the partial sums unbounded.

For the positive sequences in the final request, the inequality makes $y_n$ decreasing and bounded below by zero. The bounded monotone sequence theorem gives $y_n\to l\ge 0$. Summing the decrements givesSuppose $l>0$. Since $y_n\ge l$, this impliesBut clipping preserves divergence of a positive series: if infinitely many $x_n\ge l$, the clipped series has infinitely many terms equal to $l$; otherwise its tail equals the divergent tail of $\sum x_n$. Either case contradicts the displayed bound. Hence the only possible limit isThis proves the minimum-decrement convergence criterion without assuming that the divergent positive sequence $x_n$ itself tends to zero.

The ratio test states that, for a series $\sum a_n$ with nonzero terms eventually, if $|a_{n+1}/a_n|\to L<1$, the series converges absolutely; if $L>1$, including $L=\infty$, it diverges because its terms do not tend to zero. At $L=1$ the test is inconclusive: the harmonic series diverges while $\sum n^{-2}$ converges.

For $|x|<1$, the ratio test applied to $x^n/n$ gives absolute convergence when $x\ne 0$, and at $x=0$ every such term vanishes. ThusAt $x=1$ all terms are zero, so the series converges to zero. At $x=-1$, the even terms vanish and the odd terms are $-2/n$, whose partial sums tend to $-\infty$. Finally, if $|x|>1$, then $|x^n-1|/n\ge (|x{|}^n-1)/n\to \infty$, so the term test for divergence applies. Consequently

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.

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