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For real $z>0$, defineThe integrand is positive and at most $e^{-t}$, so $0<F(z)<1$ and the improper integral converges. Substituting $s=z+t$ also gives $F(z)=z{e}^z{E}_1(z)$ in terms of the exponential integral; the original representation makes positivity and the large-$z$ behavior transparent.

Apply the exact finite geometric series identity inside the integral, without assuming that an infinite series can be integrated term by term:The Gamma integral gives ${\int }_0^{\infty }{e}^{-t}{t}^{k}dt=k!$. Therefore, with $N$ retained terms,This remainder formula proves the asymptotic expansion at every fixed order: $R_N=O(z^{-N})=o(z^{-N+1})$ as $z\to +\infty$. Its sign also gives alternating bounds: an odd $N$ overestimates $F(z)$, and an even $N$ underestimates it.

At $z=10$, an independent evaluation of the integral gives $F(10)\approx 0.9156333394$. Three retained terms giveFive retained terms give $S_5(10)=0.9164$ with error bounded by $5!/10^5=0.0012$. The actual errors are about $0.004367$ and $0.000767$, respectively. The error bound is conservative, but established without relying on the numerical integral evaluation.

Nevertheless, the formal infinite series diverges at every fixed $z>0$. The magnitudes $T_k=k!/z^k$ satisfy $T_{k+1}/T_k=(k+1)/z$, so they eventually increase without bound. In particular, the terms fail the necessary zero-term condition for a convergent series. The finite remainder identity remains valid after convergence has failed; this is precisely what makes the example useful.

An analytic continuation extends a holomorphic function beyond its original domain without changing its existing values. Formally, if $U\subset V$ are domains and $f$ is holomorphic on $U$, a holomorphic function $F$ on $V$ with $F{|}_U=f$ is an analytic continuation. More locally, holomorphic functions on overlapping open sets continue one another when they agree on the overlap, allowing a chain of patches along a path. The identity theorem for holomorphic functions forces agreement throughout a connected overlap once they agree on a nonempty open set in it. It also makes an extension to a fixed connected domain unique. ^[ref]

The geometric series illustrates why a formula can reach farther than the power series that first represents it. For $z\ne 1$, its partial sum isWhen $|z|<1$, $z^{N+1}\to 0$, so the series converges to $f(z)=1/(1-z)$. The rational function $F(z)=1/(1-z)$ is holomorphic on $\mathbb{C}\setminus \{1\}$ and agrees there with the sum on the unit disc. It therefore continues $f$ to every point except its simple pole at $1$. In particular,Thus analytic continuation gives a value at $2$ without making the original geometric series converge there. Its radius of convergence remains $1$: on $|z|\ge 1$ the terms $z^n$ do not even tend to zero.

The pole at $1$ limits the original Taylor series, although most points of its convergence circle are regular for $F$. Recentring at $-1$ gives another genuinely convergent power series,At $z=-3/2$ this yields $\frac{1}{2}{\sum }_{n\ge 0}(-1/4{)}^n=2/5$, even though the original powers of $-3/2$ grow. This is the useful idea: overlapping local descriptions can represent one holomorphic function on a larger region.

Uniqueness on a connected overlap does not imply that every path leads to the same terminal germ of a holomorphic function. Continuing a complex logarithm around $0$ changes its value by $2\pi i$, giving monodromy; the return patch can carry a different branch of a multivalued function. If two patches have a disconnected overlap, agreement on one component imposes no agreement on the others. ^[ref] Continuation using the rational function $1/(1-z)$ is single-valued: any path avoiding $1$ gives the same terminal germ of a holomorphic function, even if it winds around the pole.

An arithmetic series adds the sequence terms of an arithmetic progression. Its $n$-th partial sum is $S_n={\sum }_{k=1}^n{a}_k$, with the empty sum convention $S_0=0$. For $n\ge 1$,Thus the total is the number of sequence terms multiplied by the arithmetic mean of the two endpoints. The formula also covers decreasing and constant arithmetic progressions.^[ref]

For the eight-entry example, the sum is $8(7+28)/2=140$, and the arithmetic mean of all eight entries is $17.5$. A decreasing example $11,9,7,5,3,1$ gives $6(11+1)/2=36$. Negative values cause no algebraic difficulty: $3,1,-1,-3,-5$ has sum $5(3-5)/2=-5$.

The triangular number $T_n=1+2+\cdots +n$ is the special case $a_1=d=1$, giving $n(n+1)/2$. More generally, an equal increase in each successive allocation makes the total grow as a quadratic function of the number of allocations. The eight-entry example hasIts first four entries add to $46$, its first eight to $140$, and its first twelve to $282$. The sequence terms lie on a straight line when plotted against their sequence indices, while the partial sums follow a quadratic function.

The direct divisor summatory function formula evaluates $N$ floor divisions. Its balanced Dirichlet hyperbola method formula evaluates $r=\lfloor \sqrt{N}\rfloor$ floor divisions and reuses their sum for the other strip, followed by doubling and square-overlap subtraction. Thus it uses $O(\sqrt{N})$ arithmetic operations, with big O notation, in a model that counts each division, addition and multiplication as one operation. At $N=10^6$, the division counts for these two formulas are $10^6$ and $10^3$.

For a general Dirichlet convolution, the asymmetric formula has $\lfloor A\rfloor +\lfloor B\rfloor$ outer terms; evaluating $F$ and $G$ has an additional cost. Under $AB=N$, equal thresholds minimize $A+B$, but an asymmetric split can be useful if the two partial sums have different evaluation costs. The balanced divisor summatory function gains the further reuse because its two arithmetic functions are identical.

These are arithmetic operation counts, rather than counts of elementary operations on binary digits: the integers grow with $N$, and division of longer integers costs more in that model of computational complexity theory. Computing $r$ also requires an exact integer square root, or checking and correcting a numerical square root until $r^2\le N<(r+1{)}^2$ holds.

A balanced Dirichlet hyperbola method split takes real thresholds $A=B=\sqrt{N}$. An asymmetric split takes $1\le A\le N$ and $B=N/A$. Every lattice point with $ab\le N$ then has $a\le A$ or $b\le B$, and every point in their overlap rectangle is allowed because $AB=N$. For arithmetic functions $f,g$, define their partial sums by $F(t)={\sum }_{a\le t}f(a)$ and $G(t)={\sum }_{b\le t}g(b)$, always over positive integers. The same inclusion-exclusion principle, now with weight $f(a)g(b)$, yieldsNo multiplicativity assumption is needed: this is a finite rearrangement of Dirichlet convolution. The constant-one arithmetic function in both positions recovers the divisor summatory function. For $N=12$, $A=2$, $B=6$ givesIt uses two columns and six rows, with a twelve-point overlap.

For general positive real thresholds, $AB\ge N$ is sufficient for coverage, since $a>A$ and $b>B$ would imply $ab>AB\ge N$. If $AB>N$, however, the overlap can include points above the hyperbola. The correct subtraction is then ${\sum }_{a\le A,b\le B,ab\le N}f(a)g(b)$; the product $F(A)G(B)$ is justified only when the whole integer rectangle lies below it. For example, $N=12$, $A=B=4$ covers every allowed pair, but its rectangle contains the forbidden pair $(4,4)$, so subtracting all sixteen points would give $34$ instead of $35$.

Conversely, the balanced integer thresholds $r,r$ have $r^2\le N$ and still cover every allowed pair because $(r+1{)}^2>N$. More generally, positive integer thresholds $U,V$ give coverage exactly when $(U+1)(V+1)>N$, and their overlap is a full rectangle exactly when $UV\le N$. These discrete conditions explain why the square-root identity works even when $r^2<N$.

A Banach space is a normed vector space over the real numbers or complex numbers that is complete for the metric $d(x,y)=\Vert x-y\Vert$. Explicitly, whenever a sequence $(x_n)$ satisfiesthere is an element $x$ of the same space with $\Vert x_n-x\Vert \to 0$. The hypothesis says that the approximations eventually agree to any requested accuracy; completeness says that they describe an actual object in the space. The Cauchy sequence condition can be checked before anyone knows what that object is.

This makes a Banach space a natural setting for constructing solutions by successive improvements. If corrections $v_k$ have ${\sum }_k\Vert v_k\Vert <\infty$, their partial sums form a Cauchy sequence by the triangle inequality, and completeness supplies their sum. In fact, convergence of every such absolutely summable vector series characterizes a Banach space. The contraction mapping theorem uses precisely this mechanism: each correction is smaller than the preceding one by a fixed factor.^[ref]

Every normed vector space of finite dimension is a Banach space. Infinite-dimensional examples include the absolutely summable sequence space and the l-infinity sequence space; the choice of norm remains part of the specification. A collection of increasingly accurate finite descriptions can require an infinite description for its limit, as the finitely supported sequence example below demonstrates. Completion of a normed space adds these missing limits while preserving all original distances.

The Basel problem asks for the exact sum of the reciprocal squares of the positive integers. The answer isThe integral test proves that the series converges, but does not identify this exact value. The Euler evaluation of even zeta values does: the equally spaced poles of cotangent connect reciprocal integer powers to Bernoulli numbers and $\pi$. This separates three different questions: whether a sum exists, how well a partial sum approximates it, and what its exact value is.

For example, the ten-term partial sum is about $1.549768$, while the exact value is about $1.644934$. The zeta series truncation error bounds the missing amount between $1/11$ and $1/10$; it is much larger than the first omitted term $1/121$. More terms improve the approximation, but a numerical coincidence with ${\pi }^2/6$ would not prove the identity.

In On the sums of series of reciprocals by Leonhard Euler, §2 identifies the reciprocal-square sum with one-sixth of the squared circumference of a unit-diameter circle. The translation's publication note dates the presentation to 5 December 1735 and publication to 1740.^[ref] The cotangent coefficient proof of the even zeta values gives a short modern outline, while zeta series truncation error addresses the distinct approximation question.

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 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.

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.

A series over the real numbers or complex numbers converges if and only if its tails satisfy: for every $\epsilon >0$, some $N$ makes $|{\sum }_{k=n+1}^m{a}_k|<\epsilon$ whenever $m>n\ge N$. This says exactly that its partial sums form a Cauchy sequence; sufficiency uses completeness.

The Cesàro mean of the first $n\ge 1$ terms of a sequence is their arithmetic mean. It is a cumulative average: increasing $n$ includes the whole initial segment, rather than a fixed-length sliding window. For $a_k=(-1{)}^k$, the even-length averages are zero and the odd-length averages are $-1/n$, so they tend to zero even though the sequence does not converge.

For a periodic sequence with period $p\ge 1$ and one-cycle arithmetic mean $M$, split an initial segment into full cycles and a remainder of fewer than $p$ terms. The full cycles contribute exactly $M$ per term; the remainder contributes a fixed bounded error divided by $n$. Thus the Cesàro means tend to $M$. The cycle $0,2,1$ in the bounded partial sums need not be Cesàro summable illustration has arithmetic mean one. Treating it as a sequence of partial sums gives a series with first-order Cesàro sum one; averaging a sequence and summing its terms remain different operations.

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.

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]

If a real or complex sequence converges to $a$, its Cesàro means have the same limit of a sequence. This is why Cesàro summation is a regular summation method when applied to the partial sums of a series.

Proof outline. Given $\epsilon >0$, choose $K$ so that $|a_k-a|<\epsilon$ for $k>K$. The triangle inequality gives, for $n>K$,The finite initial contribution tends to zero, and the remaining contribution is at most $\epsilon$. The converse fails: the alternating sequence $(-1{)}^n$ has Cesàro means tending to zero while its terms do not converge.

The approach of a variable mathematical object to a specified limit. For a sequence this concerns its terms; for a series it concerns its partial sums.

For a fixed positive integer $m$, if $a_n\to 0$ and the series of the displayed block sums converges, then the original series converges to the same limit of a sequence. Its partial sum at a block endpoint is exactly a partial sum of $\sum b_n$; the difference at an intermediate endpoint has at most $m-1$ terms, all tending to zero. This justification is essential when block cancellation proves convergence without absolute convergence. It does not apply without further control to blocks whose lengths grow.

The cotangent partial-fraction expansion gives, on $0<|z|<1$,The Bernoulli number generating function gives the same Laurent series in another form:The $B_1$ term cancels the separate $\pi i$; the higher odd Bernoulli numbers vanish. Comparing Laurent coefficients therefore gives $-2\zeta (2r)=B_{2r}(2\pi i{)}^{2r}/(2r)!$, which is the stated real formula because $i^{2r}=(-1{)}^r$. The cotangent partial-fraction expansion justifies the convergent expansion before the coefficient comparison; a finite plot of partial sums alone would not prove the infinite identity.

The cycle lemma turns a constraint at every prefix into a choice of where to start reading a circular sequence. Think of a lattice path whose vertical increments record votes for two candidates: $+1$ for one candidate and $-1$ for the other. Its partial sums are the running lead. Rotating the increments preserves the final lead, but changes whether that lead stays strictly positive throughout. This makes rotation useful for counting lattice paths subject to prefix constraints; the sum-one cycle lemma also supplies the cyclic prefix condition in the cyclic Turán covering construction.

The sum-one form. Let $b=(b_1,\dots ,b_n)$ be any nonempty sequence of integers with ${\sum }_{i=1}^n{b}_i=1$. A starting position $r\in \{0,\dots ,n-1\}$ means reading the rotated sequenceCall $r$ good when every nonempty partial sum of $b^{(r)}$ is strictly positive, including the full sum:The cycle lemma says there is exactly one good starting position. Negative entries, zero entries and positive entries larger than one are all allowed in this sum-one form. The empty partial sum is zero and is excluded from the positivity requirement. Douglas West's exercise 1.3.46(a) states a stronger result for arbitrary integer entries: take its case with all $n$ partial sums positive. ^[ref]

A worked sum-one example. For $b=(2,-2,0,1)$, the partial sums starting at zero are $S_0,S_1,S_2,S_3,S_4=0,2,0,0,1$. Starts $0,1,2$ fail: they have, respectively, a zero second partial sum, a negative first partial sum, and a zero first partial sum. Start $3$ reads $(1,2,-2,0)$ and has nonempty partial sums $1,3,1,1$, all positive. This example both allows an increment larger than one and shows why touching zero fails.

Why the last minimum works. Put $S_0=0$ and $S_j={\sum }_{i=1}^j{b}_i$. Choose the last index $r$ attaining the minimum of the proper partial sums $S_0,\dots ,S_{n-1}$: include $S_0$ and exclude $S_n$. After that index, later proper partial sums are strictly larger than $S_r$; also $S_n=1>S_r$, since $S_r\le 0$. Before wrapping, the new partial sums are therefore positive differences $S_j-S_r$. After wrapping, they are $1+S_j-S_r\ge 1$. In the worked example the last minimum is at $r=3$; choosing the first minimum at $r=0$ would fail. Uniqueness takes one further observation: two good starts would split the circle into two nonempty arcs, each having a positive integer sum, yet their sum would be one.

A standard positive-total form of the Dvoretzky-Motzkin cycle lemma uses a nonempty sequence $a=(a_1,\dots ,a_n)$ of integers satisfyingExactly $k$ starting positions have every nonempty partial sum strictly positive, using the same rotation convention as the sum-one cycle lemma. Negative entries can be arbitrarily large in magnitude, and zero entries are allowed. Addario-Berry's proposition 2.2.1 gives an equivalent form with increments at least $-1$ and negative total; reversing the sequence and changing all signs gives the form here. ^[ref]

Four starts, two successes. For $a=(1,1,1,-1)$, the sum is $k=2$. Starts $0$ and $1$ give partial sums $(1,2,3,2)$ and $(1,2,1,2)$, respectively. Start $2$ gives $(1,0,1,2)$ and touches zero; start $3$ gives $(-1,0,1,2)$ and begins below zero. Thus exactly two starts are good. The rotated sequence $(-1,1,1,1)$ has the same positive total, but its first start is bad: a positive final lead alone does not control earlier partial sums.

Positions can give identical words. The sequence $(1,1,-1,1,1,-1)$ repeats a block of length three and has sum two. Its good starts are $0$ and $3$, both giving the same word $(1,1,-1,1,1,-1)$ with partial sums $(1,2,1,2,3,2)$. Starts $1$ and $4$ touch zero; starts $2$ and $5$ begin negatively. The cycle lemma with increments at most one counts two starting positions even though there is only one distinct good rotated word. Counting distinct words would lose the multiplicity that makes the statement true.

The upper bound matters. The one-entry sequence $(2)$ has positive integer sum two, but only one starting position. Thus the sum-$k$ conclusion fails for arbitrary positive increments when $k>1$. The sum-one cycle lemma above needs no such upper bound. Strict positivity also matters: counting paths that merely stay at or above zero would include additional starts in both diagrams.

From rotation to enumeration. For a fixed admissible sequence, a start chosen from a uniform distribution on a finite set of its $n$ positions is good with probability $k/n$. More generally, let $\mathcal{F}$ be any finite set of length-$n$ sequences satisfying the increment bound and total $k$, closed under rotation. Double counting pairs consisting of a sequence and a good starting position givesFor each good output sequence and each of the $n$ positions, inverse rotation gives one such pair, even when several rotated words coincide. For example, the four sequences containing three $+1$ votes and one $-1$ vote have exactly $\frac{2}{4}\left(\genfrac{}{}{0ex}{}{4}{3}\right)=2$ vote orders with a strictly positive running lead: $(1,1,1,-1)$ and $(1,1,-1,1)$. This is how the cycle lemma with increments at most one converts many prefix inequalities into a simple counting factor.