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For a positive random variable $X$, let $D\in \{1,\dots ,9\}$ be its first nonzero base-ten digit. The Benford law is the discrete probability distribution with probability mass functionHere $X$ need not be an integer: $0.0042$, $4.2$ and $4200$ all have $D=4$. Multiplication by an integer power of ten changes the decimal point's position and leaves the first nonzero digit unchanged. Zero has no first nonzero digit, so exclude it. For negative inputs one may explicitly use their absolute values, still excluding zero; that convention must be stated rather than applying a logarithm to a negative number.

The Benford law favors small digits: $\mathbb{P}(D=1)={\log }_{10}2\approx 0.301029996$, whereas $\mathbb{P}(D=9)={\log }_{10}(10/9)\approx 0.04575749$. Thus the model assigns about $30.10\%$ to digit one and $4.58\%$ to digit nine. The masses are positive and their normalization is a finite telescoping series:

Why can digit one be so common? In a decimal significand interval $[1,10)$, the digit-one region $[1,2)$ occupies a larger fraction of the logarithmic scale than the digit-nine region $[9,10)$. The precise assumption is that the fractional part of the base-ten logarithm is distributed uniformly on $[0,1)$; Benford law from logarithmic uniformity turns those unequal interval lengths into probabilities. This makes a surprising digit pattern into a simple geometry of multiplicative size. The logarithmic explanation is also presented in Peter Sheridan Dodds, Benford's Law, essential story.

The Benford law is useful as a proposed model for leading digits when the mechanism supports logarithmic uniformity or an appropriate limiting argument, such as Benford frequencies for powers of an integer. Merely spanning many powers of ten does not establish that assumption. A discrepancy from the Benford law can prompt investigation of the model and data collection; by itself it does not establish fraud, and agreement does not establish authenticity.

Simon Newcomb's 1881 paper, Note on the Frequency of Use of the Different Digits in Natural Numbers, appeared in the American Journal of Mathematics 4, pp. 39–40. Its opening connects unequal leading-digit frequencies with the faster wear of early pages in logarithm tables. The publication metadata and original opening excerpt are available in Newcomb (1881), original publication record and ^[ref]. Frank Benford's The Law of Anomalous Numbers, Proceedings of the American Philosophical Society 78 (1938), pp. 551–572, likewise discusses worn logarithm tables on p. 551 and tabulates observations on p. 553. See Benford (1938), transcription of the original paper with page numbers and publication metadata. These observations motivated study of the digit pattern; they do not make the Benford law universal.

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 finite discrete random variables, the mutual information supplied by two observations equals the information supplied by the first plus the additional conditional mutual information supplied by the second:The order changes the individual contributions but leaves the total unchanged.^[ref] To prove the first identity, add and subtract the intermediate conditional entropy:With base-two logarithms, all contributions are in bits. Repeatedly inserting intermediate conditional entropies makes them cancel, as in a telescoping series, and gives the rule for any finite list of observations:where the $j=1$ term is $I(X;Y_1)$. The same calculation with $W$ already known gives $I(X;Y,Z\mid W)=I(X;Y\mid W)+I(X;Z\mid Y,W)$.

For the fair-bit exclusive or example in conditional mutual information, observing $Y$ alone tells nothing about $X$, but observing $Z$ after $Y$ tells everything:Here $H(X\mid Y)=1$ by independence and $H(X\mid Y,Z)=0$ because $X=Y\oplus Z$. In the reverse order, $H(X\mid Z)=1$, so $I(X;Z)=0$ and $I(X;Y\mid Z)=1$: the two observations are individually uninformative but jointly reveal one bit. Simply adding their separate mutual informations would miss all of it.

The chain rule is useful for measuring the value of each new observation without double counting information already supplied. It also guarantees $I(X;Y,Z)\ge I(X;Y)$ by nonnegativity of conditional mutual information. Keeping $Y$ and adding $Z$ can help, even though discarding $Y$ and retaining only a processed $Z$ is constrained by the data processing inequality. Equating the two observation orders gives the precise identitywhich explains why there is no general rule that conditioning decreases mutual information.

For $x_{n+1}=a_n{x}_n+b_n$, put $U_0=1$ and $U_n={\prod }_{j=1}^n{a}_j$, with $a_j\ne 0$. Dividing by this accumulated homogeneous growth gives $z_n=x_n/U_{n-1}$ and $z_{n+1}-z_n=b_n/U_n$. A telescoping series then provesThis is the discrete counterpart of the integrating factor for a first-order linear differential equation.

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.

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

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

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

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

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

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

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

The key telescoping series identity isFor the requested mathematical induction, when $n=1$ the proposed sum is $q/(q+1)=1-q!/(q+1)!$. If the identity holds for $n$, adding $c_{n+1}=q!/(q+n)!-q!/(q+n+1)!$ cancels its last term and yields $1-q!/(q+n+1)!$. Thus it holds for every $n\ge 1$. Since $q!/(q+n)!\to 0$ for fixed positive integer $q$, the infinite series has sum $\boxed{1}$.

For the second series, the integral digit condition is $0\le a_n\le n-1$, and in particular $a_1=0$. Its partial sums are increasing and satisfyThus the bounded monotone sequence theorem proves convergence to a real number $S\in [0,1]$. The same telescoping calculation gives, for every $N\ge 1$,If positive digits occur infinitely often, then every tail contains a positive term, so $R_N>0$. If $a_n\le n-2$ occurs infinitely often, then every tail contains a digit smaller than its maximum. Indeed,Consequently the two conditions together give $0<R_N<1$ for every $N$. Were $S=A/B$ a rational number, with positive integer $B$, choose $N\ge B$. Because $B\mid N!$, the number $N!S$ is an integer, as is ${\sum }_{n=1}^N{a}_{n}N!/n!$. Their difference $R_N$ would be an integer strictly between zero and one, a contradiction.

Conversely, if positive digits occur only finitely often, then $S$ is a finite sum of rational numbers. If digits below their maximum occur only finitely often, then $a_n=n-1$ for every $n>N$ for some $N$, givingThese are exactly the two ways the conjunction of infinitude conditions can fail. Hence the irrationality criterion for factorial series is proved in both directions: $S$ is irrational exactly when both kinds of digit occur infinitely often.

Put $N=\lfloor x\rfloor$. Comparing the harmonic number with an integral gives $H_N={\sum }_{n=1}^{N}1/n>{\int }_1^{N+1}dt/t=\log (N+1)>\log x$. By the fundamental theorem of arithmetic, the finite Euler product ${\prod }_{p\le x}(1-p^{-1}{)}^{-1}$ expands as the sum of $1/n$ over positive integers whose prime factors are at most $x$. It therefore includes every term of $H_N$, andThe first strict inequality is the geometric-series bound for the logarithmic remainder; the last uses the telescoping series ${\sum }_{n=2}^{\infty }1/[n(n-1)]=1$. Thus the prime reciprocal lower bound is

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.

Subtract successive terms and use a telescoping series:Thus the linear recurrence relation has solution

Put $U_0=1$, so the same definition $z_n=x_n/U_{n-1}$ includes $n=1$. Since $U_n=a_n{U}_{n-1}$ and every $a_n$ is nonzero,The factor $U_n^{-1}$ is a discrete integrating factor for this variable-coefficient affine recurrence. Sum the increments as a telescoping series and multiply back by $U_n$:The identity requires no assumption that the coefficients have the same sign.

Expand the finite Euler product into geometric series. By unique prime factorization, it is the sum of $1/n$ over positive integers whose prime factors do not exceed $x$, and hence contains every term up to $m=\lfloor x\rfloor$. The harmonic sum satisfiesThe last inequality uses $m+1>x$, including when $x$ is an integer. For each prime number $p$,The telescoping series over all integers $n\ge 2$ bounds the sum of these remainders by $\frac{1}{2}{\sum }_{n\ge 2}[n(n-1){]}^{-1}=1/2$. Taking logarithms of the product bound therefore givesThis explicit constant suffices; no asymptotic theorem about the distribution of prime numbers is needed.

The first series is telescoping, sinceIts $N$th partial sum is $1-1/(N+1)$, soa rational number.

The second series converges absolutely, for example by the ratio test, and is the odd part of the exponential series:This value is irrational. Indeed, if $a=\sinh 1$ were algebraic, then $e$ would satisfymaking $e$ algebraic over the algebraic numbers and hence algebraic, contrary to the Hermite theorem on the transcendence of e. Thus the second sum is in fact transcendental.

The telescoping series has partial sum $S_N=w_{N+1}-w_1$. Therefore the necessary and sufficient condition is that $w_n$ have a finite limit $L$. The sum is then $L-w_1$.

Put $p=k-1>0$. The preceding inequality bounds $n^{-k}<[(n-1{)}^{-p}-n^{-p}]/p$ for $n\ge 2$. For $N\ge 2$, summing the telescoping series givesThese nonnegative partial sums increase and are bounded, so the monotone convergence theorem for sequences proves convergence for every $k>1$. This is the telescoping bound for a p-series, an alternative to the integral test.

Write $\rho =|\mathbf{x}-\mathbf{y}|$. A rotationally symmetric fundamental solution of the Laplace equation is harmonic for $\rho >0$. The Laplace equation in polar coordinates reduces to $(\rho G^{\prime }{)}^{\prime }=0$, so $G=A\log \rho +B$. Integrate $-\Delta G={\delta }^{(2)}(\mathbf{x}-\mathbf{y})$ over a disk of radius $ϵ$ around the source. The divergence theorem givesTaking $B=0$ therefore givesChanging the reference length inside the logarithm merely adds a constant. Notice the sign convention: the potential equation in this question is $\Delta V=q{\delta }^{(2)}$, whereas this Green function satisfies $-\Delta G={\delta }^{(2)}$. Thus the point-source potential used below is $-qG=+(q/2\pi )\log \rho$.

For the two insulating walls, successive Euclidean reflections in $x=b$ and $x=-b$ generate sources of strength $q$ at $(2nb,0)$ for every integer $n$. Only $n=0$ lies inside the physical strip. This is the Neumann image lattice for a point source in a strip. SetThe raw sum of logarithmic potentials diverges by additive constants. A well-defined potential is obtained by pairing opposite images and subtracting their constant reference terms:Each paired summand is $O(n^{-2})$ on compact subsets away from the source. The series therefore has local uniform convergence, as do the required derivative series. It givesIn the corresponding $x$-derivative, the images pair as $n$ and $1-n$ at $x=b$, and as $n$ and $-1-n$ at $x=-b$. Their opposite horizontal separations make the derivatives cancel. Thus both homogeneous Neumann boundary conditions hold. The constructed solution is even in $y$; without a condition at infinity the side conditions alone would allow further homogeneous harmonic functions.

For fixed $y>0$, uniform convergence on $[-b,b]$ permits integration term by term. The $n$th term has integralHere $\arctan$ is the ordinary real inverse trigonometric function, taking values in $(-\pi /2,\pi /2)$. Summing from $-N$ to $N$ makes a telescoping series:The source flux splits equally between the upper and lower parts of the strip. The next two parts give the single-wall constructions that underlie the image lattice.

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.

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.

For $p>0$ and $n\ge 2$, integrating the decreasing derivative magnitude of $x^{-p}$ gives $(n-1{)}^{-p}-n^{-p}=p{\int }_{n-1}^n{x}^{-p-1}dx>p{n}^{-p-1}$. Put $p=k-1$ and sum this telescoping series to obtain the bound in the title, strict for $N>1$. The partial sums are increasing and bounded, proving convergence of the p-series when $k>1$. This gives a direct comparison test for series proof and a finite upper bound without evaluating the sum.