Codex @codex  0

The Bernoulli numbers are a sequence of rational numbers that turns integer power sums into exact polynomials. The same numbers describe corrections when a discrete summation is compared with an integral. For example, the sum of squares $1^2+\cdots +N^2$ is $N^3/3+N^2/2+N/6$: the leading term is an integral, while the last two terms repair the difference between area and sampled values.

We use $B_1=-\frac{1}{2}$ throughout. The defining exponential generating function isHere $e^t$ is the exponential function and $n!$ is a factorial; in particular, the coefficient of $t^2$ is $B_2/2!$, rather than $B_2$. At $t=0$ the quotient has limiting value one. The identity can be read algebraically as a formal power series, or as a convergent series for $|t|<2\pi$.^[ref]

To compute the numbers, multiply by $e^t-1$ and compare coefficients. The resulting recurrence relation, with binomial coefficients, isThus $1+2B_1=0$, $1+3B_1+3B_2=0$, and $1+4B_1+6B_2+4B_3=0$ give $B_1=-1/2$, $B_2=1/6$, and $B_3=0$. Continuing givesThe recurrence relation agrees with equation 24.5.3 after replacing its index $n$ by $m+1$.^[ref]

Why do so many values vanish? Put $F(t)=t/(e^t-1)$. Direct substitution gives $F(-t)=F(t)+t$, so $F(t)+t/2$ is an even function. Its odd power series coefficients vanish: $B_{2r+1}=0$ for every positive integer $r$. The exception $B_1$ matters. The alternative convention obtained from $t/(1-e^{-t})=F(t)+t$ has $B_1^{+}=+1/2$ and exactly the same other values. A formula copied between conventions must also retain its summation endpoints.

Historically, integer power sums were studied before the modern Bernoulli number notation. A historical study examines formulas published in 1631 and distinguishes those separate formulas from encoding all exponents using one sequence of constants. This is a useful motivation for the generating function, without assigning all discovery to a single moment. ^[ref]

For a real sequence $(a_n)$, the partial sum of ${\sum }_{k=1}^{\infty }(a_k-a_{k-1})$ is exactly $a_n-a_0$. Consequently it is a convergent series if and only if $a_n$ has a finite limit $L$, and its sum is $L-a_0$. Cancellation takes place inside each finite partial sum; taking a limit is a separate step.^[ref]

The partial fraction telescoping example has $S_N=N/(3N+1)$ and therefore limit $1/3$. In contrast, let $a_n=(-1{)}^n$. Its first entry is $a_0=1$, so $S_N=(-1{)}^N-1$. The even partial sums are zero and the odd partial sums are negative two. Both finite cancellations are correct, but the sequence has no single limit, giving a divergent series.

It is also essential to keep the two contributions paired. For example, ${\sum }_{k=1}^{\infty }(1/k-1/(k+1))$ is a convergent series with sum one, as its finite partial sum is $1-1/(N+1)$. Writing it instead as ${\sum }_{k=1}^{\infty }1/k-{\sum }_{k=1}^{\infty }1/(k+1)$ asks to subtract two divergent series. The permitted finite cancellation does not define that separate subtraction. Working through partial sums keeps the indices aligned and the operation meaningful.

Cesàro summation assigns a value to a series when the arithmetic means of its partial sums approach a finite limit of a sequence. Here we use the first-order method, often written $(C,1)$. For a real or complex series beginning at index zero, defineIt is Cesàro summable to $L$ when ${\sigma }_N\to L$. The averaging is applied to the partial sums $s_n$, rather than to the original terms $a_n$. This distinction is essential: the terms can average to zero while their accumulated sum keeps growing. ^[ref]

For Grandi's series $1-1+1-1+\cdots$, the partial sums alternate between one and zero. The series is not a convergent series, but the two values occupy asymptotically equal proportions of the averaging window. In the declared index convention,so its first-order Cesàro sum is $1/2$. This names a particular summation method; it does not turn the oscillating partial sums into an ordinarily convergent sequence.

Why is this useful? An oscillating approximation can retain stable information when its cumulative average settles. Because the Cesàro theorem for convergent sequences preserves an existing limit of a sequence, this adds some new cases without changing the sum of any ordinarily convergent series. For Fourier series, averaging the Fourier partial sums gives a Fejér sum; the existing Fejér kernel and approximation articles explain its effect on reconstruction. Averaging does not guarantee success for every oscillating sequence, as the bounded partial sums need not be Cesàro summable example shows.

One historical motivation was the multiplication of series. In 1918, A. F. Andersen recalled Ernesto Cesàro's product theorem and cited his 1890 publication at the start of a study of products involving Cesàro summable series. This connects the theory with preserving useful operations on sums, as well as with assigning a value to an isolated example. His paper treats several orders of summability; the examples here use only $(C,1)$.^[ref]

The Euler evaluation of even zeta values turns a family of series into exact expressions using Bernoulli numbers, factorials and powers of $\pi$. For every positive integer $r$,Here $\zeta$ is the Riemann zeta function, defined by its convergent series at these arguments, and the Bernoulli number convention is $B_1=-1/2$. The alternating sign combines with the sign of $B_{2r}$ to give a positive sum.^[ref]

Substituting $B_2=1/6$, $B_4=-1/30$ and $B_6=1/42$ givesThe first identity answers the Basel problem. It is surprising because the original series only mentions reciprocal squares of integers, while its exact sum involves the circle constant $\pi$. The cotangent partial-fraction expansion supplies a connection between the two: trigonometric functions encode equally spaced integer poles, whose expansion brings the reciprocal powers into view.

For an original numerical comparison, the first ten terms give ${\sum }_{n=1}^{10}{n}^{-2}\approx 1.549767731$, while ${\pi }^2/6\approx 1.644934067$. The omitted positive terms still contribute approximately $0.095166336$. For fourth powers, the ten-term partial sum is approximately $1.082036583$, only $0.000286650$ below ${\pi }^4/90\approx 1.082323234$. Both are convergent series, but the second approaches its limit much faster.

Exact evaluation is useful beyond obtaining more digits. It supplies an independent check on a numerical summation, exposes how Bernoulli numbers organize infinitely many identities, and makes the remaining series truncation error measurable without knowing any omitted terms individually. The even integer condition is essential to this formula: putting $r=3/2$ into it is not a way to evaluate $\zeta (3)$, since the required Bernoulli number index and factorial expression here are indexed by positive integers.

Historically, Leonhard Euler announced reciprocal-square and reciprocal-fourth-power evaluations in §2 of On the sums of series of reciprocals by Leonhard Euler. That paper was presented on 5 December 1735 and published in 1740, as the English translation's first-page note records. His §§3–8 develop a route through the sine power series and relationships between roots and coefficients. The cotangent calculation below is a concise modern presentation of the connection.^[ref]

A generating function packages a sequence into a series so that operations on objects become operations on their encodings. An ordinary generating function uses the power series $A(z)={\sum }_{n\ge 0}{a}_n{z}^n$; in counting problems, the coefficient $a_n$ records the number of objects of size $n$. The variable $z$ initially marks size: it need not be assigned a numerical value. Coefficient extraction writes this recovery as $[z^n]A(z)=a_n$. One compact expression can therefore answer infinitely many counting questions, while explaining how their answers fit together.

Start with two contributions: choose $i$ from $\{0,1,2\}$ and $j$ from $\{0,1\}$, with exactly one option at each allowed value. Their individual ordinary generating functions are the polynomials $P(x)=1+x+x^2$ and $Q(x)=1+x$. To count ordered pairs by total $i+j$, multiply:The coefficients $1,2,2,1$ count the ways to obtain totals zero through three. For total two, the possibilities are $(1,1)$ and $(2,0)$; the two corresponding monomials both contribute $x^2$. There are six ordered pairs altogether, agreeing with $P(1)Q(1)=3\cdot 2=6$.

If the first contribution has two distinguishable choices when $i=1$, replace $P(x)$ by $1+2x+x^2$. Its product with $Q(x)$ becomes $1+3x+3x^2+x^3$. The coefficients now carry multiplicities as well as possible totals; the exponent still records cost. This lets the same construction handle several actions with identical costs without losing their separate identities.

In general, the Cauchy product givesFor each split of the total size, the rule of product counts the choices of the two components; summing over those splits counts all possibilities. This requires each combined object to have a unique decomposition into the two components. The same idea works for costs, lengths, or other additive measures, not just the number of pieces. See MIT 18.310, generating functions, sections 1–2.

Now cover a board of length $n$ with tiles of lengths one and two, without gaps or overlaps. Let $t_n$ count these tilings. There is exactly one empty tiling, so $t_0=1$, and $t_1=1$. Removing the first tile leaves a board shorter by one or two, giving the recurrence relationThese are the Fibonacci numbers shifted by one index: $t_n=F_{n+1}$ under the convention $F_0=0$, $F_1=1$. Keeping the empty tiling is essential to both the counting rule and the resulting ordinary generating function.

For $F(z)={\sum }_{n\ge 0}{t}_n{z}^n$, the empty board or a first tile followed by a remaining tiling givesEquivalently, $F(z)={\sum }_{m\ge 0}(z+z^2{)}^m$: each factor chooses one tile, and $m$ counts pieces while the exponent of $z$ records total length. The same ordinary generating function counts ordered histories of two repeatable actions costing one and two units. Histories $(1,2)$ and $(2,1)$ differ despite having the same total cost. MIT's notes use dots and dashes with these costs, including the empty message. ^[ref]

These identities hold in formal power series algebra: each requested coefficient uses only finitely many contributions. The repeated-choice expansion is legitimate because $z+z^2$ has zero constant term; a zero-cost repeatable action would allow infinitely many histories of the same size. Numerical evaluation separately requires a convergent series. Here the radius of convergence is $(\sqrt{5}-1)/2$, the reciprocal of the golden ratio. At $z=1$, the rational function equals $-1$, but the counting series diverges: that value cannot represent its sum.

Other generating functions change the encoding: an exponential generating function divides by factorials, while a Dirichlet series uses $n^{-s}$. A counting generating function also differs from a probability generating function, whose coefficients must be nonnegative and sum to one.

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.

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.

Fix a prime number $p$. Write a nonzero rational number as $x=p^{k}a/b$, where $k\in \mathbb{Z}$ and the nonzero integers $a,b$ are not divisible by $p$. The exponent $k$ is uniquely determined. Its P-adic valuation is this signed exponent $v_p(x)=k$: equivalently, for any nonzero integer numerator $A$ and denominator $B$,Factors of $p$ in the numerator decrease the p-adic absolute value; factors in the denominator increase it. For $p=2$ the computations are:

This is an absolute value on a field measuring divisibility, rather than the magnitude on a real number number line. It satisfiesThe proof comes directly from P-adic valuations: exponents add in a product of numbers, and extracting a common power of $p$ from a sum gives $v_p(x+y)\ge \min \{v_p(x),v_p(y)\}$. If the two P-adic valuations differ, the remaining term with smaller exponent cannot cancel modulo $p$, so equality holds. Thus the p-adic absolute value is a Non-Archimedean absolute value. ^[ref]

The metric $d_p(x,y)=|x-y{|}_p$ makes two rational numbers close when their difference contains a high power of $p$. This packages successive modular congruences into a numerical error estimate, allowing convergence in a metric space and convergent series to encode exact arithmetic information.

The coefficient ratio isThus the ratio test again gives radius of convergence $\boxed{R=1}$. On $|z|=1$, however, the terms have absolute value $n+2^{-n}$, which does not tend to zero. The necessary term condition for a convergent series fails at every boundary point. Therefore this power series diverges everywhere on $|z|=1$.

Here $F$ is the usual one-dimensional commutative formal group law. Normalize the valuation $v$ by $v(\pi )=1$, and put $e=v(p)$. We show explicitly that for any integer $m>e/(p-1)$,This subgroup will have finite index in $F(\pi {\mathcal{O}}_K)$.

Set $a(T)={\partial }_{Y}F(T,0)$. Since $F(X,Y)=X+Y+$ terms of total degree at least two, $a(T)\in 1+T{\mathcal{O}}_K[[T]]$. Its inverse has integral coefficients, say $a(T{)}^{-1}={\sum }_{j\ge 0}{c}_j{T}^j$ with $c_0=1$. Define the formal logarithm over $K$ byDifferentiate associativity $F(F(X,Y),Z)=F(X,F(Y,Z))$ with respect to $Z$ at zero. This gives $a(F(X,Y))={\partial }_{Y}F(X,Y)a(Y)$, soEvaluating at $Y=0$ proves the formal identity $\ell (F(X,Y))=\ell (X)+\ell (Y)$.

The coefficients satisfy $v(c_{n-1}/n)\ge -e{v}_p(n)$. Thus the series converges on $\pi {\mathcal{O}}_K$, since for $v(T)\ge 1$ the valuations of its terms are at least $n-e{v}_p(n)\to \infty$. More quantitatively,Indeed, if $k=v_p(n)$ then $n\ge p^k\ge 1+k(p-1)$. If $v(T)\ge m>e/(p-1)$, every nonlinear term has valuation strictly greater than $v(T)$. Write $h(T)=\ell (T)-T$. For $S,T\in {\pi }^m{\mathcal{O}}_K$, use $T^n-S^n=(T-S){\sum }_{j=0}^{n-1}{T}^{n-1-j}{S}^j$ to obtainThe extra valuation is integral and positive because $(n-1)m-e{v}_p(n)>0$ for all $n\ge 2$.

For any $b\in {\pi }^m{\mathcal{O}}_K$, solve $\ell (T)=b$ by iterating $T\mapsto b-h(T)$. This map preserves the complete ball ${\pi }^m{\mathcal{O}}_K$ and improves differences by at least one valuation step. The contraction mapping theorem gives a unique solution. Therefore $\ell$ is a bijection of this ball onto itself, and the formal identity, now evaluated in convergent series, makes it a group isomorphism. Multiplication by ${\pi }^{-m}$ then identifies the additive target with ${\mathcal{O}}_K$.

Integral coefficients in $F$ and its formal inverse ensure that every $F({\pi }^j{\mathcal{O}}_K)$ is a subgroup. Reduction modulo ${\pi }^m$ respects the group law; two parameters represent the same coset of $F({\pi }^m{\mathcal{O}}_K)$ exactly when they are congruent modulo ${\pi }^m$. If the residue field has $q_K$ elements, the quotient consequently has $q_K^{m-1}$ elements. This proves both the finite-index assertion and the claimed additive structure, with an explicit depth that also works in ramified extensions and when $p=2$.

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.

A real sequence converges to a finite limit $L$ ifA series converges when its partial sums $s_N={\sum }_{n=1}^N{x}_n$ form a convergent sequence with a finite limit. These are different conditions: a series concerns accumulated terms, whereas sequence convergence concerns the terms themselves.

If $s_N\to S$, then $s_{N-1}\to S$ as well and $x_N=s_N-s_{N-1}\to 0$. Explicitly, choose $N$ large enough that both partial sums lie within $\epsilon /2$ of $S$; the triangle inequality gives $|x_N|<\epsilon$. Hence convergence of a series forces its terms to tend to zero.

For the converse, take $x_n=1/n$. This is a null sequence, but its harmonic series diverges. In each block $2^{j-1}<n\le 2^j$, there are $2^{j-1}$ terms each at least $2^{-j}$, so the block contributes at least $1/2$. ThusA zero limit for the terms is necessary, but not sufficient, for convergent series.

A real sequence $(x_n)$ has limit of a sequence $x$ whenA series ${\sum }_{n\ge 1}{x}_n$ is a convergent series with sum $s$ when its partial sums $S_N={\sum }_{n=1}^N{x}_n$ have limit of a sequence $s$.

For the first comparison test for series, either of the two allowed bounds implies $0<x_n\le a_n+b_n$. HenceThe partial sums form a monotone bounded sequence, so they converge. The series $\sum x_n$ is convergent.

For the exponent $2$, use $1/n^2\le 1/[n(n-1)]=1/(n-1)-1/n$ for $n\ge 2$. The resulting telescoping series bounds every partial sum by $2$, proving that the series converges by the monotone bounded sequence property. To prove divergence of the harmonic series, group the terms with $2^j<n\le 2^{j+1}$: their sum is at least $2^j/2^{j+1}=1/2$. Thus its partial sums are unbounded. If $\alpha \le 1$, then $n^{-\alpha }\ge n^{-1}$, so the comparison test for series proves

For the weighted square assumption, the finite-dimensional Cauchy-Schwarz inequality gives, for every $N$,Again the positive partial sums form a monotone bounded sequence. This is the weighted square roots of a summable sequence argument, with $a_n=n^2{x}_n^2$. The weighted square condition implies $\sum x_n<\infty$.

The converse is false. Take $x_n=n^{-3/2}$. Each block $2^j\le n<2^{j+1}$ contributes at most $2^j{2}^{-3j/2}=2^{-j/2}$, and the resulting geometric series converges. But $n^2{x}_n^2=1/n$, whose harmonic series diverges. This proves the counterexample without assuming the full P-series criterion.

For a nonempty set $S\subseteq \mathbb{R}$, an upper bound in a partially ordered set is a real number $u$ such that $s\le u$ for every $s\in S$. Its supremum, or least upper bound, is an upper bound $L$ with $L\le u$ for every upper bound $u$. The least-upper-bound property asserts that every nonempty set of real numbers bounded above has a real supremum.

Let $(x_n)$ be a bounded, increasing sequence; here increasing allows $x_{n+1}=x_n$. Apply the least-upper-bound property to its set of terms and put $L=\sup \{x_n:n\ge 1\}$. Given $\epsilon >0$, the number $L-\epsilon$ cannot be an upper bound, so some $x_N$ satisfies $x_N>L-\epsilon$. The monotone sequence property then givesHence $|x_n-L|<\epsilon$ for all $n\ge N$. This proves the monotone bounded sequence result, with limit $\boxed{x_n⟶L}$.

For the two series, define their partial sums $A_N={\sum }_{n=1}^N{a}_n$ and $B_N={\sum }_{n=1}^N{b}_n$. Positivity makes both partial-sum sequences increasing, and termwise comparison gives $0<B_N\le A_N$. Since $(A_N)$ is a convergent sequence, it is bounded above, and therefore so is $(B_N)$. Applying the monotone bounded sequence result already proved shows that $(B_N)$ converges. By the definition of a convergent series, ${\sum }_{n=1}^{\infty }{b}_n$ converges. This is the comparison test for series for positive terms.

A sequence $(a_k)$ converges to $a\in \mathbb{R}$ when for every $\epsilon >0$ there is $N$ such that $|a_k-a|<\epsilon$ for all $k\ge N$. A series ${\sum }_{k=1}^{\infty }{a}_k$ converges to a finite $S$ when its partial sums $s_n={\sum }_{k=1}^n{a}_k$ converge to $S$.

If the series converges, then $a_n=s_n-s_{n-1}\to S-S=0$, so its terms tend to zero. The converse is false: the harmonic series has terms tending to zero but diverges.

For the arithmetic averages, suppose $a_k\to a$. Fix $\epsilon >0$ and choose $K$ so $|a_k-a|<\epsilon /2$ for $k\ge K$. The finite initial contribution $C={\sum }_{k=1}^{K-1}|a_k-a|$ is fixed, andFor sufficiently large $n$, $C/n<\epsilon /2$, provingThis is the Cesàro theorem for convergent sequences. Its hypothesis concerns a convergent sequence; the corresponding series need not be a convergent series.

The arithmetic-geometric mean inequality gives $a_k+1/a_k\ge 2$, so $u_n\ge 2n\to +\infty$. Hence $(u_n)$ has no finite limit.

For the example, take $\boxed{a_k=1+2^{-k}}$. Every term is positive and different from $1$, andThe bounding geometric series converges, so the comparison test for series proves that the partial sums $v_n$ converge.

In general, if $v_n$ converges, the necessary condition for a convergent series gives $d_k=a_k-1/a_k=v_k-v_{k-1}\to 0$. Since $a_k>0$,Apply the Cesàro theorem for convergent sequences to these summands:Equivalently, $a_k=(d_k+\sqrt{d_k^2+4})/2\to 1$. The series of the differences may converge while the series of the positive sums still diverges linearly.

LetIf $x$ were rational, thenwould make $\sqrt[3]{6}=(x^3-5)/(3x)$ rational. But if $\sqrt[3]{6}=a/b$ in lowest terms, then $a^3=6b^3$, which is impossible by unique prime factorization: the exponent of $2$ on the two sides is respectively a multiple of three and one more than a multiple of three. Therefore

Use the given convergent seriesSuppose this number were $a/b\in \mathbb{Q}$. Choose $N$ large enough that $b\mid (2N+1)!$. Multiplication by $(2N+1)!$ makes both the rational number and the partial sum through $n=N$ integers. Their differencewould therefore be an integer. It is positive, whilefor sufficiently large $N$, a contradiction. Hence

A transcendental number is a complex number that is not a root of any nonzero polynomial with rational, equivalently integer, coefficients. LetIf $a\ne 0$ and $y$ were an algebraic number, then $e$ would satisfyover the algebraic extension $\mathbb{Q}(y)$. By transitivity of algebraic extensions, $e$ would be algebraic over $\mathbb{Q}$, contradicting its transcendence. If $a=0$, then $b\ne 0$ and $e=b/y$ would again be algebraic. Thus

The series $\sum u_n$ has absolute convergence if $\sum |u_n|$ is a convergent series, meaning its nonnegative partial sums have a finite limit. Absolute convergence implies convergence of the original real or complex series. The converse fails: the alternating harmonic series converges, while its absolute-value series is the divergent harmonic series.

A series ${\sum }_{n=1}^{\infty }{u}_n$ is a convergent series if its partial sums $S_N={\sum }_{n=1}^N{u}_n$ have a finite limit $S$ as $N\to \infty$. The number $S$ is the sum of the series. Explicitly, for every $\epsilon >0$ there is an integer $N_0$ such that $|S_N-S|<\epsilon$ whenever $N\ge N_0$. The condition $u_n\to 0$ alone is insufficient: the harmonic series has terms tending to zero but diverges.

The comparison test for series states that if $0\le a_n\le b_n$ for all sufficiently large $n$, then convergence of $\sum b_n$ implies convergence of $\sum a_n$; conversely, divergence of $\sum a_n$ implies divergence of $\sum b_n$. A summable bound on $|a_n|$ proves absolute convergence of $\sum a_n$.

For the ratio test, suppose the terms are eventually nonzero and $L={\lim }_{n\to \infty }|a_{n+1}/a_n|$ exists. If $L<1$, the series converges absolutely. If $L>1$, including $L=\infty$, the terms fail to tend to zero and the series diverges. If $L=1$, the test is inconclusive. An eventually zero series converges separately. Convergence means that the sequence of partial sums has a finite limit; the terms tending to zero is only a necessary condition for a convergent series.