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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.

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]

The Bernoulli polynomials extend the Bernoulli numbers by inserting the exponential function $e^{xt}$ into their exponential generating function:For example, $B_0(x)=1$, $B_1(x)=x-1/2$, $B_2(x)=x^2-x+1/6$, and $B_3(x)=x^3-3x^2/2+x/2$.^[ref]

Their crucial property is a finite difference identity:Indeed, subtracting the two exponential generating functions cancels $e^t-1$ and leaves $t{e}^{xt}$. Thus $B_{p+1}(x)/(p+1)$ plays for discrete summation the role that $x^{p+1}/(p+1)$ plays for integration: taking a unit finite difference recovers $x^p$.^[ref]

Consider $N$ noninteracting bosons in one internal quantum state, at thermal equilibrium in the three-dimensional harmonic oscillator potentialHere $m$ is the mass, the positive ${\omega }_i$ are angular frequencies, and $\bar{\omega }$ is their geometric mean. Measure the one-particle energy $ϵ$ and chemical potential $\mu$ from the ground-state energy. The quantum harmonic oscillator levels then have $ϵ=\hslash ({\omega }_x{n}_x+{\omega }_y{n}_y+{\omega }_z{n}_z)$ with nonnegative integer $n_i$ and reduced Planck constant $\hslash =h/(2\pi )$.

When $ϵ\gg \hslash {\omega }_i$, counting lattice points can be approximated by the volume of a tetrahedron in the positive octant of $(n_x,n_y,n_z)$:This is a density of states, with units of inverse energy. The factor $1/6$ is the unit tetrahedron volume; differentiating produces the factor $1/2$. Treat the single ground state separately: its macroscopic population would be lost by replacing every level with a smooth density of states.

For $k_{B}T\gg \hslash {\omega }_i$, integrate the Bose-Einstein distribution over excited energy eigenstates. As $\mu$ approaches zero from below, their maximum population isHere $k_B$ is the Boltzmann constant and $\zeta$ the Riemann zeta function. A short proof uses $1/(e^x-1)={\sum }_{\ell \ge 1}{e}^{-\ell x}$ and ${\int }_0^{\infty }{x}^2{e}^{-\ell x}dx=2/{\ell }^3$. The nonnegative terms permit exchanging summation and integration by Tonelli's theorem, giving ${\int }_0^{\infty }{x}^2/(e^x-1)dx=2\zeta (3)$. Near zero the integrand is proportional to $x$, so its integral converges. That finite excited-state capacity is the mechanism of Bose-Einstein condensation.

Set $N_{\mathrm{ex}}^{\mathrm{sat}}(T_c)=N$. Particle conservation gives, to this continuum accuracy,Above $T_c$, the macroscopic condensate fraction is zero and the chemical potential adjusts below zero. This cubic power belongs to a three-dimensional harmonic trap; it is not the homogeneous three-dimensional ideal gas exponent. For $N=10^5$, $\bar{\omega }/(2\pi )=100\mathrm{Hz}$ and $\zeta (3)\simeq 1.202056903$, the model gives $T_c\simeq 209.5\mathrm{nK}$. At $T=0.8T_c\simeq 167.6\mathrm{nK}$, $N_0/N=0.488$, or approximately $48800$ condensed bosons. These are chosen parameters and calculated values.

Finite $N$ gives a smooth crossover rather than a sharp phase transition. The exact ideal gas sum over excited quantum harmonic oscillator levels differs from the continuum integral, especially when $k_{B}T$ is not large relative to every $\hslash {\omega }_i$. A sharp trapped-gas limit takes $N\to \infty$, $\bar{\omega }\to 0$ with $N{\bar{\omega }}^3$ fixed and fixed frequency ratios. Interactions also alter the critical temperature, condensate fraction and release profile; a dilute gas can have appreciable interaction energy compared with the trap level spacing. The cubic law is a useful baseline, not a fit guaranteed for either 1995 experiment.

Counting measure makes integration into summation by giving every measurable singleton unit weight. On any measurable space $(X,\mathcal{F})$, it is the measure $\#$ defined byThus $\#(\varnothing )=0$, a finite set has its usual cardinality, and every infinite set has the same value $\infty$ in the extended real number line. In particular, countably infinite sets and uncountable sets are not distinguished by their counting measure. Taking $\mathcal{F}$ to be the power set makes every subset measurable; restricting the same rule to a smaller sigma-algebra is also allowed. ^[ref]

Why is this a countably additive measure? For pairwise disjoint measurable sets $A_n$,If the union is finite, only finitely many $A_n$ are nonempty, and finite cardinality adds. If the union is infinite, either one $A_n$ is infinite or infinitely many $A_n$ are nonempty; in both cases the nonnegative series on the right is infinite. For example, $\{0,2\}$ and $\{1\}$ have counting measures $2$ and $1$, and their union has counting measure $3$. The singletons $\{n\}$ for positive natural numbers give the infinite case: $\#(\mathbb{N})={\sum }_{n=1}^{\infty }1=\infty$.

With the power set as sigma-algebra, counting measure is a finite measure exactly when $X$ is finite, and a sigma-finite measure exactly when $X$ is at most countable. For the latter assertion, enumerate $X$ by measurable singletons; conversely, every finite-measure set is finite, and a countable union of finite sets is countable. Specifying the sigma-algebra matters: if $X=\mathbb{N}$ and $\mathcal{F}=\{\varnothing ,\mathbb{N}\}$, counting measure is not a sigma-finite measure, since the only measurable finite-measure set is the empty set.

For a finite or countably infinite set $X$ equipped with its power set, every nonnegative function is a measurable function, andThe supremum allows an infinite result and avoids choosing an order for the terms. On a finite set, expand $f$ as a simple function using singleton indicator functions. On a countably infinite set, enumerate the points and apply the monotone convergence theorem to finite truncations. Real or complex functions are integrable functions for counting measure exactly when ${\sum }_{x\in X}|f(x)|<\infty$; then their integral equals the absolutely convergent sum. This is the discrete counterpart of Lebesgue integration, developed further in series as a Lebesgue integral against counting measure. ^[ref]

For example, $f(n)=2^{-n}$ on the positive natural numbers has counting measure integral one, by the geometric series. The alternating harmonic series has a finite ordered sum but is not the integral of an integrable function for counting measure: its positive and negative parts both have infinite sums. Absolute convergence is what makes the finite integral independent of enumeration. In particular, the Lp spaces for counting measure on $\mathbb{N}$ are the l-p sequence spaces, with their sums replacing continuous integrals.

Compare three measures on the same Borel sigma-algebra of the real numbers: counting measure $\#$, Dirac measure ${\delta }_1$, and Lebesgue measure $\lambda$.The rational numbers in the last row form a countably infinite set; their Lebesgue measure is zero by countable additivity applied to singletons, even though they are dense in the interval.

These are different answers to different questions: counting measure asks how many points occur, Dirac measure asks whether the selected point occurs, and Lebesgue measure asks how much length occurs. A null set therefore depends on its measure. For counting measure, the only null set is the empty set, so equality almost everywhere is equality everywhere. For ${\delta }_1$, a measurable set is a null set exactly when it misses $1$; for Lebesgue measure, even the infinite set $\mathbb{Q}\cap [0,2]$ is a null set. Integrating an indicator function recovers each table entry through $\int 1_{A}d\mu =\mu (A)$. This lets the same Lebesgue integral framework handle point counts, point evaluation, and geometric length.

The Dirichlet hyperbola method turns a long summation over a Dirichlet convolution into two short ones by covering the lattice points below a hyperbola with two overlapping strips. Its simplest application is an exact count: for a positive integer $N$, every positive integer divisor $a$ of $n\le N$ corresponds to the ordered pair $(a,n/a)$. Thus the divisor summatory function counts precisely the ordered pairs of positive integers with $ab\le N$:Here $\tau$ is the divisor function, and the floor function gives the number of allowed values of $b$ for each $a$. The axes are excluded, while points on $ab=N$ are included.

Set $r=\lfloor \sqrt{N}\rfloor$, using the floor function and the positive square root. Every counted pair has $a\le r$ or $b\le r$. Otherwise $ab\ge (r+1{)}^2>N$. The first strip contains ${\sum }_{a=1}^r\lfloor N/a\rfloor$ points; swapping the two coordinates gives the same count for the second. Their overlap contains all $r^2$ pairs with $1\le a,b\le r$, since $r^2\le N$. The inclusion-exclusion principle therefore gives the exact identityThe subtraction is for the whole overlap square, rather than only the diagonal. The same Dirichlet hyperbola method identity holds for real $x\ge 1$, with $D(x)={\sum }_{n\le x}\tau (n)$ and $r=\lfloor \sqrt{x}\rfloor$. ^[ref]

A discrete uniform distribution gives every element of a nonempty finite set $S$ the same probability. Write $n=|S|$ for its cardinality. A random variable $X$ with this law has support of a probability distribution $S$ and probability mass functionThe masses sum to $n(1/n)=1$. Conversely, if all $n$ masses equal $p$, their sum is $np=1$, so necessarily $p=1/n$. Thus equal chances determine the whole law.^[ref]

Counting now computes an event's probability: for any set $A$ of possible values,In particular, when $A\subseteq S$, the answer is $|A|/|S|$. The intersection matters: values outside the support of a probability distribution contribute zero. This turns a probability question into a combinatorics question, provided that equal chances are justified by the model.

For an ideal fair six-sided die, $S=\{1,2,3,4,5,6\}$ and each face has probability $1/6$. The event of an even integer is $\{2,4,6\}$, so its probability is $3/6=1/2$; the event $X\ge 5$ has probability $2/6=1/3$. Direct summation gives the expected value $\mathbb{E}X=21/6=7/2$ and the variance $\mathrm{Var}(X)=91/6-(7/2{)}^2=35/12$. The mean $7/2$ is a balance point between possible outcomes. These are exact consequences of stipulated fairness, rather than measured frequencies from physical dice.

Uniformity describes probabilities, not distances between values. For a second constructed model, stipulate equal-chance selection from $S=\{1,2,4,8\}$. Every value has probability $1/4$, despite the unequal gaps $1,2,4$. Here $\mathbb{P}(X\ge 4)=2/4=1/2$, while the expected value and variance areNeither the midpoint of the extreme values, $9/2$, nor a formula using only the cardinality gives these moments. For a general numerical support of a probability distribution, use its actual values:

Choosing an item from a finite list with equal chances is a basic building block of simple random sampling. A bijection that relabels the items preserves equal probabilities, although changing numerical labels can change the expected value and variance. A function that merges labels can destroy uniformity: for the fair-die model, $Y=\min (X,3)$ has masses $1/6,1/6,4/6$ at $1,2,3$, since four faces map to three. Always count the original equally likely outcomes.

An indicator random variable for membership in a set $A\subseteq S$ has a Bernoulli distribution with parameter $|A|/n$; counting that event over $r$ independent repetitions gives a binomial distribution with parameters $r$ and $|A|/n$. Equal chances among individual values need not produce equal chances among aggregated counts.^[ref]

The divisor summatory function adds the divisor function up to a cutoff $x\ge 1$. It counts all ordered pairs of positive integers whose product is at most $x$. For $N=12$, the twelve individual divisor function values areCounting columns of lattice points instead gives $12+6+4+3+2+2+1+1+1+1+1+1=35$. The Dirichlet hyperbola method needs only the first three column counts:The arithmetic work of the Dirichlet hyperbola method is much smaller even though the answer is identical.

This is useful for elementary averages: individual divisor function values fluctuate, but their summation has the regular behavior given by the divisor summatory asymptotic. Dividing that estimate by a positive integer $N$ giveswhere $\log$ is the natural logarithm, $\gamma$ is the Euler--Mascheroni constant, and $O$ uses big O notation. This describes an average, rather than an estimate for every individual divisor function value. ^[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]

For an arithmetic function $a$, changing the order of the finite summation gives ${\sum }_{d=1}^{n}a(d)\lfloor n/d\rfloor ={\sum }_{t=1}^n{\sum }_{d\le \lfloor n/t\rfloor }a(d)$. Both sides count pairs of positive integers $(d,t)$ with $dt\le n$, assigning weight $a(d)$ to each pair. This expresses a weighted floor sum in terms of a summatory function.

A summation algorithm that keeps a correction for low-order information lost when adding floating-point numbers. Feeding that correction into subsequent additions can greatly reduce accumulated rounding error compared with straightforward summation.

For integrable real-valued functions $f,g$ on a measure space $(X,\mu )$ and fixed real numbers $a,b$, their linear combination $af+bg$ is integrable, and the Lebesgue integral satisfiesThe same rule applies to any finite summation: ${\int }_X{\sum }_{j=1}^n{a}_j{f}_{j}d\mu ={\sum }_{j=1}^n{a}_j{\int }_X{f}_{j}d\mu$ for integrable functions $f_j$ and fixed scalars $a_j$. Finite signed values matter: integrability ensures all these integrals are finite, so no undefined $\infty -\infty$ occurs.

This is the distributive property of finite summation carried into integration: a fixed factor can be pulled out, and a finite sum can be integrated term by term. For simple functions, distribute the finite sums of value times measure; approximation extends additivity to nonnegative measurable functions, and splitting into a positive part and negative part gives the integrable signed case. See Lecture 12: Lebesgue Integrable Functions, the Lebesgue Integral and the Dominated Convergence Theorem (Proposition 122: linearity for integrable functions). For example, ${\int }_0^1(2x-3)dx=2{\int }_0^{1}xdx-3{\int }_0^{1}1dx=-2$. With a probability measure, linearity of integration gives linearity of expectation.

The two finite summations cancel term by term after shifting the index. Only the final positive term remains:This is a telescoping sum.

A prime sum is a summation indexed by prime numbers, for example ${\sum }_{p\le x}1/p$. The indexing restriction distinguishes it from a sum over all positive integers.

Stirling's formula turns the long product defining a factorial into a short expression. For positive integers $n$, putThen $n!\sim S_n$ as $n\to \infty$. This is asymptotic equivalence: $S_n/n!\to 1$, so the relative error tends to zero. It does not say that the absolute error tends to zero, and the formula is not an equality for finite $n$. The Gamma function extends this approximation beyond integer inputs. ^[ref]

For $n=10$,The leading estimate is about $0.830\%$ low. Multiplying by the first correction factor giveswhich is about $0.003176\%$ low. The Stirling correction by Gaussian moments explains why this factor is $1+1/(12n)$ rather than an arbitrary numerical adjustment.

The usefulness is in ratios as well as in enormous numbers. Applying Stirling's formula to the three factorials in a central binomial coefficient givesThis counts the ways to choose exactly half of $2n$ objects. Equivalently, for $2n$ independent fair coin tosses, the binomial distribution assigns probability $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n\sim 1/\sqrt{\pi n}$ to exactly $n$ heads. An exponentially large count becomes a simple probability scale after cancellation.

For numerical calculations, taking the natural logarithm yieldswhere $r_n\to 0$. Computing a logarithm of a large factorial avoids representing the factorial itself. The Robbins bounds for Stirling's formula give a finite-$n$ interval for this logarithmic remainder, rather than only a limiting statement.

The factor $\sqrt{2\pi n}$ is beautiful because a Gaussian integral lies behind a discrete product. The Gaussian peak behind Stirling's formula displays this connection through the Gamma integral and Laplace's method. A rigorous proof must also control the tails of the integral; the local expansion explains the geometry but is not by itself a complete proof.

James Stirling published Methodus differentialis in 1730, a treatise on summation and interpolation of infinite series. This setting connects the formula with a practical calculation problem: taking logarithms turns a long product into a sum, whose growth can be approximated. ^[ref]

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 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.

A triangular number counts rows of $1,2,\dots ,n$ objects, with $n$ a nonnegative integer:The first values are $0,1,3,6,10,15,21,28,36,45,\dots$. The empty arrangement contributes zero, by the empty sum convention. Adding one more row gives the triangular number recurrence $T_n-T_{n-1}=n$ for $n\ge 1$. The row lengths form an arithmetic progression, while their running totals are triangular numbers.

Pair the rows with a reversed copy: $1+n=2+(n-1)=\cdots =n+1$. There are $n$ such pairs, so two copies contain $n(n+1)$ dots. Dividing by two proves the formula. This rectangle picture explains the factor one-half without requiring a formula for a long summation. For example, one hundred rows contain $T_{100}=5050$ dots.

The same count appears whenever every pair of different objects must be considered once. Five participants make ten possible handshakes; five sensors have ten different pairwise comparisons; a complete graph on five vertices has ten edges. These are the triangular number pair-counting identity, with one fewer row than objects. The distinction prevents a common error: $T_5=15$ counts all pairs from six objects, whereas five objects give $T_4=10$.

The triangular number square-complement identity says that two neighboring triangular numbers fill a square. Combining that identity with the one-row difference makes the difference of squares $T_n^2-T_{n-1}^2$ equal to $n^3$. It leads to Nicomachus's theorem: the sum of the first $n$ cubes of numbers is the square of a number $T_n^2$. A change in the way one groups objects exposes a relation between summation, multiplication and geometry.

This geometric view is already visible in Nicomachus's Introduction to Arithmetic by Nicomachus (Hahn teacher edition): Book II, Chapter 8 describes growing triangular rows, Chapter 12 combines successive triangular counts into squares, and Chapter 20 groups successive odd integers into cubes of numbers.^[ref] The linked edition uses Martin Luther D'Ooge's translation, revised with modern teaching material by Nathanael Hahn. Our empty-row index $T_0=0$ is a modern convention; the historical passage starts from unity and calls three the first actual triangle.