Series (mathematics, ${\sum }_{n=1}^{\infty }{a}_n$)

A series adds the terms of a sequence in their specified order. Its infinite-sum notation is defined through finite partial sums:when the displayed limit of a sequence exists as a finite number. Otherwise the series is not a convergent series in the ordinary sense. Writing an infinite list of additions does not by itself assign it a finite sum. Cesàro summation is a different prescription, with its own definition.

Two positive series make the distinction concrete. For $a_n=1/n$, the individual terms tend to zero but the harmonic series has unbounded partial sums. For $a_n=1/[n(n+1)]$, partial fraction decomposition reveals a telescoping series whose partial sums tend to one. The useful question is how much all the remaining terms can contribute together, rather than only how small the next term looks.

A series whose terms are multiples of sines and cosines of integer multiples of a variable. A Fourier series is the special case with coefficients extracted from a specified function by integration.

A Lambert series has the form ${\sum }_{n\ge 1}{a}_n{q}^n/(1-q^n)$. Expanding each denominator as a geometric series shows that its coefficient of $q^m$ is ${\sum }_{d\mid m}{a}_d$.

A double series is a series whose terms carry two indices. For nonnegative terms, define its sum as the supremum of all finite subsums; iterated sums in either order give this same value, possibly infinity. For absolutely summable terms, both orders also agree and give a finite sum. Without these hypotheses, an order of summation must be specified and changing it can change the result. A triangular index set, such as $0\le n\le m$, can be represented by setting the other terms to zero.

A series whose partial sums do not have a finite limit of a sequence. They may grow without bound or oscillate. Some divergent series belong to the domain of a summation method, such as Cesàro summation.

A rule for assigning a value to some series through a specified limiting operation. It must say which series belong to its domain; different rules can have different domains. A regular summation method agrees with ordinary summation for every convergent series.

A summation method that includes every ordinarily convergent series and assigns it its ordinary sum. The Cesàro theorem for convergent sequences makes first-order Cesàro summation regular. Regularity alone does not justify every rearrangement or multiplication of a divergent series.

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]

In the average of partial sums $s_0,\dots ,s_N$, the term $a_k$ occurs exactly $N-k+1$ times. Reordering this finite sum gives the displayed identity. Early terms have weights closer to one, and the newest term has weight $1/(N+1)$. The triangular weight expresses the cumulative averaging rule; it is not a claim that one may freely rearrange an infinite series.

The series with terms $a_n=(-1{)}^n$, indexed from zero, named for Guido Grandi. Its even-indexed partial sums are one and its odd-indexed partial sums are zero, so it has no ordinary sum. First-order Cesàro summation gives $1/2$. Grouping adjacent pairs would obscure this distinction: the partial sums of the grouped expressions select only one of the original subsequences.

For the geometric series with terms $z^n$, $z\ne 1$, the partial sums are $(1-z^{n+1})/(1-z)$. Averaging the finite expressions givesIf $|z|\le 1$ and $z\ne 1$, the second term has absolute value at most $2/((N+1)|1-z{|}^2)$, and therefore tends to zero. On the unit circle, except at $z=1$, this yields a first-order Cesàro sum even though the terms do not tend to zero. For $z=1$, instead $s_n=n+1$ and ${\sigma }_N=(N+2)/2$, so there is no finite Cesàro sum.^[ref]

The finite limit of a sequence of averaged partial sums in the declared Cesàro summation convention, when that limit exists. It can exist when an ordinary sum does not, but every ordinarily convergent series has the same Cesàro sum as its ordinary sum.

Construct partial sums $s_n$ from alternating blocks of zero and one, with lengths $1,3,9,27,\dots$. Block $j$, beginning with $j=0$, has length $3^j$ and value zero when $j$ is even, one when $j$ is odd. Define a series by $a_0=s_0$ and $a_n=s_n-s_{n-1}$ for $n\ge 1$; the telescoping series identity recovers these chosen partial sums. They remain in $[0,1]$.

At the end of odd block $2m+1$, the Cesàro mean is exactly $3/4$. At the end of even block $2m$, it isThese two subsequences have different subsequential limits, so the series has no first-order Cesàro sum. A later block remains a substantial fraction of everything averaged so far. Boundedness alone therefore does not ensure a stable average.

For $a_n=(-1{)}^n(n+1)$, direct finite summation gives $s_{2m}=m+1$ and $s_{2m+1}=-(m+1)$. ConsequentlyThe distinct even and odd subsequential limits prevent first-order Cesàro summation. More averaging is a different method; a value assigned by another method is not a first-order Cesàro sum.

A factorial series with digits $0\le a_n<n$ is an absolutely convergent series of real numbers between zero and one. The telescoping series identityshows that its largest possible tail after place $N$ is $1/N!$. Multiplying this tail by $N!$ gives a number between zero and one, a useful arithmetic constraint on rational numbers.

A factorial series with integral digits $0\le a_n<n$ is an irrational number exactly when positive digits occur infinitely often and digits below their maximum occur infinitely often. Under these two conditions every scaled tail $N!{\sum }_{n>N}{a}_n/n!$ lies strictly between zero and one. If the sum were a rational number $A/B$, choose $N\ge B$; then $N!A/B$ and the scaled finite head are integers, contradicting the strict tail bound. Conversely, an eventually zero digit sequence gives a finite rational number sum, and an eventually maximal sequence gives a finite head plus $1/N!$, again a rational number.

For $p,q>0$, the positive series converges precisely when $p>1$, or $p=1$ and $q>1$. For $p>1$ use comparison with a convergent p-series; for $0<p<1$, the bound $(\log n{)}^q<n^{(1-p)/2}$ gives a lower comparison with a divergent p-series. At $p=1$, the integral test for convergence and $u=\log x$ reduce the question to $\int u^{-q}du$. In contrast, ${\sum }_{n\ge 3}[n(\log \log n{)}^r{]}^{-1}$ diverges for every finite $r>0$, since $(\log \log n{)}^r<\log n$ eventually and $\sum [n\log n{]}^{-1}$ diverges.

For a series ${\sum }_n{a}_n$, let $r={\mathrm{limsup}}_{n\to \infty }|a_n{|}^{1/n}$. If $r<1$, choose $r<\rho <1$; eventually $|a_n|\le {\rho }^n$, so geometric comparison gives absolute convergence. If $r>1$, the terms fail to tend to zero along a subsequence and the series diverges. At $r=1$, the test is inconclusive.

A telescoping series has partial sums in which internal contributions cancel. The basic pattern is $a_n=b_n-b_{n+1}$, for which finite summation givesOnly the two boundary terms survive. If $b_n$ tends to a finite limit of a sequence $L$, the series consequently sums to $b_1-L$. The cancellation is exact for each finite $N$; the infinite result then follows by a limit of a sequence.

For example, partial fraction decomposition givesThe first three terms are $1/2,1/6,1/12$, with partial sum $3/4$. The first five have partial sum $5/6$. Taking the limit of a sequence gives the concise answer

The appeal is that a long calculation becomes a boundary calculation. It works whenever the summand is a successive difference, not only for rational functions. For instance, $a_n=\log ((n+1)/n)=\log (n+1)-\log n$ gives partial sum $\log (N+1)$, so its series diverges despite the cancellation. A telescoping series therefore need not be a convergent series: one must still examine the surviving boundary term.

For the telescoping series ${\sum }_{n=1}^{\infty }1/[n(n+1)]=1$, the remainder after the first $N$ terms is exactlyThus $S_9=0.9$, $S_{99}=0.99$, and $S_{999}=0.999$. This supplies an exact stopping rule: to make the absolute error at most $10^{-3}$, 999 terms suffice. Requiring strictly less than $10^{-3}$ instead needs at least 1000 terms.

The next term, $a_{N+1}=1/[(N+1)(N+2)]$, is smaller than the full remaining sum by a factor of $N+2$. For example, after 99 terms the next term is $1/10100$, while the remainder is $1/100$. The term test for divergence concerns individual terms; a reliable numerical approximation needs a bound on the whole remaining sum.

The term test for divergence says that a series cannot converge if its terms fail to tend to zero. Indeed, if its partial sums satisfy $S_n\to S$, then $a_n=S_n-S_{n-1}\to S-S=0$. This is a necessary condition for a convergent series.

The converse fails: $1/n\to 0$, but the harmonic series diverges. For a quick successful test, ${\sum }_{n=1}^{\infty }n/(n+1)$ diverges because its terms tend to one. The test gives no conclusion for $\sum 1/n$ or $\sum 1/[n(n+1)]$: both have terms tending to zero, yet only the second is a convergent series. The telescoping series calculation and the harmonic series block estimate resolve those cases.

An infinite product is the limit of the finite partial products ${\prod }_{n=1}^N{a}_n$. If $a_n=1+u_n$, every factor is nonzero, and ${\sum }_n|u_n|<\infty$, then the product converges to a nonzero value.

For $|q|>1$, define $S_q(z)={\prod }_{k\ge 0}(1-q^{-2k-1}z)(1-q^{-2k-1}{z}^{-1})$ on ${\mathbb{C}}^{\ast }$. The geometric decay gives locally uniform absolute tail convergence there. Its simple zeros are $q^{2k+1}$ and $q^{-2k-1}$ for $k\ge 0$, and reindexing gives $S_q(qz)=-z{S}_q(q^{-1}z)$. The variable zero is outside its domain, and the poles introduced by each reciprocal factor cannot be interpreted as a value of the product at zero.

// Destination: analysis.bigb

The Weierstrass elementary factor has a simple zero at $w=1$ and no other zeros. Its logarithmic series begins at degree $p+1$: $\log E_p(w)=-{\sum }_{\nu \ge p+1}{w}^{\nu }/\nu$ for $|w|<1$. Raising the factor order $p$ makes the tail arbitrarily small on a fixed compact set and enables infinite product convergence from logarithmic tails for unrestricted discrete zero data.

If a tail of holomorphic factors admits logarithms whose absolute values have a summable bound uniformly on every compact set, their product converges locally uniformly to the exponential of the logarithmic sum. The tail is holomorphic and nowhere zero. Finite initial factors may have zeros; those finite factors determine the locations and orders of all zeros near any given compact set. The scalar sufficient condition ${\sum }_j|u_j|<\infty$ for factors $1+u_j$ follows from $|\log (1+u_j)|\le 2|u_j|$ for $|u_j|\le 1/2$.

For $a_j\ge 0$, the finite product satisfiesThe lower inequality follows by product expansion; the upper one uses $1+t\le e^t$. Since both the partial sums and partial products are increasing, one converges to a finite limit exactly when the other does. The monotone bounded sequence theorem proves both implications. The product cannot tend to zero in this nonnegative setting.

This infinite product converges uniformly on compact subsets of the complex plane. Pair the factors at $iz$ and $-iz$ in the Weierstrass product for the reciprocal gamma function, and use the Gamma reflection formula together with $\Gamma (1+w)=w\Gamma (w)$. This gives $[\Gamma (1+iz)\Gamma (1-iz){]}^{-1}=\sinh (\pi z)/(\pi z)$ and the displayed product. The removable value at $z=0$ is one. Matching zeros alone would not exclude multiplication by a nonvanishing entire function; the normalized product fixes that ambiguity.

A lattice sum is a series indexed by the points of a lattice. For example, comparison with annular lattice-point counts shows that ${\sum }_{(m,n)\ne (0,0)}(m^2+n^2{)}^{-s}$ converges when $s>1$.

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.

For a convergent series with sum $S$ and $N$-th partial sum $S_N$, its series remainder is $R_N=S-S_N$. It equals the sum of the terms after position $N$. Its absolute value measures the error of replacing $S$ by $S_N$; it need not be positive when the series has signed sequence terms.

Let $(b_n)$ be positive and increase to infinity. If the series ${\sum }_{n\ge 1}{x}_n/b_n$ converges, thenThis follows from summation by parts: write $x_k/b_k$ as a convergent series and average its tails with the increasing weights $b_k$.

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 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.

If $(a_n)$ is a nonnegative decreasing sequence, then ${\sum }_{n\ge 1}{a}_n$ converges if and only if the condensed series ${\sum }_{k\ge 0}{2}^k{a}_{2^k}$ converges. Comparing each dyadic block with its first and last terms proves both directions.