User:Twoxili/sandbox

List of representations of e

As a recursive function

The series representation of ${\displaystyle e}$ , given as $${\displaystyle e={\frac {1}{0!}}+{\frac {1}{1!}}+{\frac {1}{2!}}+{\frac {1}{3!}}+\cdots }$$ can also be expressed using a form of recursion. When ${\displaystyle \textstyle {\frac {1}{n}}}$  is iteratively factored from the original series the result is the nested series^[1] $${\displaystyle e=1+{\frac {1}{1}}\left(1+{\frac {1}{2}}\left(1+{\frac {1}{3}}\left(1+\cdots \right)\right)\right)}$$ which equates to $${\displaystyle e=1+{\cfrac {1+{\cfrac {1+{\cfrac {1+\cdots }{3}}}{2}}}{1}}}$$  This fraction is of the form ${\displaystyle \textstyle f(n)=1+{\frac {f(n+1)}{n}}}$ , where ${\displaystyle f(1)}$  computes the sum of the terms from ${\displaystyle 1}$  to ${\displaystyle \infty }$ .

As a binomial series

Consider the sequence:

${\displaystyle e_{n}=\left(1+{\frac {1}{n}}\right)^{n}}$ 

By the binomial theorem^[2]:

${\displaystyle e_{n}=\sum _{k=0}^{n}{n \choose k}{\frac {1}{n^{k}}}=\sum _{k=0}^{n}{\frac {n^{\underline {k}}}{k!}}{\frac {1}{n^{k}}}}$ 

which converges to ${\displaystyle e}$  as ${\displaystyle n}$  increases. The term ${\displaystyle n^{\underline {k}}}$  is the ${\displaystyle k}$ th falling factorial power of ${\displaystyle n}$ , which behaves like ${\displaystyle n^{k}}$  when ${\displaystyle n}$  is large. For fixed ${\displaystyle k}$  and as ${\displaystyle \textstyle n\to \infty }$ :

${\displaystyle {\frac {n^{\underline {k}}}{n^{k}}}\approx 1-{\frac {k(k-1)}{2n}}}$ 

As a product of integrals

The definite integral of ${\displaystyle \textstyle f(x)}$  over the interval ${\displaystyle \textstyle [0,1]}$  can be approximated by the limit of a Riemann sum:

${\displaystyle \int _{0}^{1}f(x)\,dx=\lim _{k\to \infty }\sum _{i=1}^{k}f(x_{i})\Delta x}$ 

For the function ${\displaystyle \textstyle f(x)=2x}$  over the interval ${\displaystyle \textstyle [0,1]}$ , the Riemann sum with partition width ${\displaystyle \textstyle \Delta x={\frac {1}{k}}}$  is given as^[3][4][5]:

${\displaystyle U_{f,P_{k}}=\sum _{i=1}^{k}\left(2\cdot {\frac {i}{k}}\cdot {\frac {1}{k}}\right)={\frac {2}{k^{2}}}\sum _{i=1}^{k}i={\frac {k(k+1)}{k^{2}}}=1+{\frac {1}{k}}}$ 

As ${\displaystyle k}$  approaches infinity, the term ${\displaystyle \textstyle {\frac {1}{k}}}$  approaches zero:

${\displaystyle 1+{\frac {1}{k}}\to 1}$ 

However, the sum ${\displaystyle \textstyle 1+{\frac {1}{k}}}$  is slightly greater than 1 for any finite ${\displaystyle k}$  before taking the limit. In the following limit:

${\displaystyle \lim _{k\to \infty }\left(\int _{0}^{1}2x\,dx\right)^{k}}$ 

if we directly evaluate the integral as 1:

${\displaystyle \int _{0}^{1}2x\,dx=\left[x^{2}\right]_{0}^{1}=1^{2}-0^{2}=1}$ 

and raise it to the power ${\displaystyle k}$ , we obtain the precise result:

${\displaystyle \lim _{k\to \infty }1^{k}=e^{0}}$ 

Instead, if we consider the Riemann sum approximation ${\displaystyle \textstyle U_{f,P_{k}}=1+{\frac {1}{k}}}$  and raise it to the power ${\displaystyle k}$ , we have the product:

${\displaystyle U_{f}^{k}=\lim _{k\to \infty }\left(1+{\frac {1}{k}}\right)^{k}=e}$ 

or ${\displaystyle \textstyle L_{f}^{k}=e^{-1}}$  for ${\displaystyle \textstyle L_{f,P_{k}}=1+{\frac {-1}{k}}}$ , with 1 as the mean. The Riemann sum approximation, when raised to ${\displaystyle k}$  (the number of partitions), converges to powers of ${\displaystyle e}$  in the limit due to the effects of continuous compounding—where small increments precipitate exponential growth over an increasing number of partitions. The result also reflects the balancing interaction between the infinitesimal deviation ${\displaystyle \textstyle {\frac {1}{k}}}$  (which represents the difference between the theoretical value of the integral and the approximation) and the exponentiation by ${\displaystyle k}$ , which prevents this difference from vanishing in the limit.

As a ratio of ratios

A unique representation of e can be found within the structure of Pascal's Triangle, as discovered by Harlan Brothers. Pascal's Triangle is composed of binomial coefficients, which are traditionally summed to derive polynomial expansions. However, Brothers identified a product-based relationship between these coefficients that links to e. Specifically, the ratio of the products of binomial coefficients in adjacent rows of Pascal's Triangle tends to e as the row number n increases:

${\displaystyle {\begin{aligned}P(n)&=\sum _{k=0}^{n}\ln {\binom {n}{k}}\\A&=P(n-1),B=P(n),C=P(n+1)\\x&=(A-B)+(C-B)\sim 1\\\end{aligned}}}$ 

${\displaystyle \exp x\sim e}$ 

The details of this relationship and its proof are outlined in the discussion on the properties of the rows of Pascal's Triangle.^[6][7]

Euler's Constant

Properties

Relation to the zeta function

The constant ${\displaystyle \gamma }$  can also be expressed in terms of the sum of the reciprocals of non-trivial zeros ${\displaystyle \rho }$  of the zeta function^[8]:

${\displaystyle \gamma =\log 4\pi +\sum _{\rho }{\frac {2}{\rho }}-2}$ 

Relation to triangular numbers

Numerous formulations have been derived that express ${\displaystyle \gamma }$  in terms of sums and logarithms of triangular numbers^{[9][10][11][12]}. One of the earliest of these is a formula^[13][14] for the ${\displaystyle n}$ th harmonic number attributed to Srinivasa Ramanujan where ${\displaystyle \gamma }$  is related to ${\displaystyle \textstyle \ln 2T_{k}}$  in a series that considers the powers of ${\displaystyle \textstyle {\frac {1}{T_{k}}}}$  (an earlier, less-generalizable proof^[15][16] by Ernesto Cesàro gives the first two terms of the series, with an error term):

${\displaystyle {\begin{aligned}\gamma &=H_{u}-{\frac {1}{2}}\ln 2T_{u}-\sum _{k=1}^{v}{\frac {R(k)}{T_{u}^{k}}}-\Theta _{v}\,{\frac {R(v+1)}{T_{u}^{v+1}}}\end{aligned}}}$ 

From Stirling's approximation^[9][17] follows a similar series:

${\displaystyle \gamma =\ln 2\pi -\sum _{k=2}^{\infty }{\frac {\zeta (k)}{T_{k}}}}$ 

The series of inverse triangular numbers also features in the study of the Basel problem^[18][19][20] posed by Pietro Mengoli. Mengoli proved that ${\displaystyle \textstyle \sum _{k=1}^{\infty }{\frac {1}{2T_{k}}}=1}$ , a result Jacob Bernoulli later used to estimate the value of ${\displaystyle \zeta (2)}$ , placing it between ${\displaystyle 1}$  and ${\displaystyle \textstyle \sum _{k=1}^{\infty }{\frac {2}{2T_{k}}}=\sum _{k=1}^{\infty }{\frac {1}{T_{k}}}=2}$ . This identity appears in a formula used by Bernhard Riemann to compute roots of the zeta function^[21], where ${\displaystyle \gamma }$  is expressed in terms of the sum of roots ${\displaystyle \rho }$  plus the difference between Boya's expansion and the series of exact unit fractions ${\displaystyle \textstyle \sum _{k=1}^{\infty }{\frac {1}{T_{k}}}}$ :

${\displaystyle \gamma -\ln 2=\ln 2\pi +\sum _{\rho }{\frac {2}{\rho }}-\sum _{k=1}^{\infty }{\frac {1}{T_{k}}}}$ 

List of logarithmic identities

Calculus identities

Integral definition

${\displaystyle \ln x=\int _{1}^{x}{\frac {1}{t}}\ dt}$ 

To modify the limits of integration to run from ${\displaystyle x}$  to ${\displaystyle 1}$ , we change the order of integration, which changes the sign of the integral:

${\displaystyle -\int _{1}^{x}{\frac {1}{t}}\,dt=\int _{x}^{1}{\frac {1}{t}}\,dt}$ 

Therefore:

${\displaystyle \ln {\frac {1}{x}}=\int _{x}^{1}{\frac {1}{t}}\,dt}$ 

Riemann Sum

${\displaystyle \ln(n+1)=}$ 

${\displaystyle \lim _{k\to \infty }\sum _{i=1}^{k}{\frac {1}{x_{i}}}\Delta x=}$ 

${\displaystyle \lim _{k\to \infty }\sum _{i=1}^{k}{\frac {1}{1+{\frac {i-1}{k}}n}}\cdot {\frac {n}{k}}=}$ 

${\displaystyle \lim _{k\to \infty }\sum _{x=1}^{k\cdot n}{\frac {1}{1+x/k}}\cdot {\frac {1}{k}}=}$ 

${\displaystyle \lim _{k\to \infty }\sum _{x=1}^{k\cdot n}{\frac {1}{k+x}}=\lim _{k\to \infty }\sum _{x=k+1}^{k\cdot n+k}{\frac {1}{x}}=\lim _{k\to \infty }\sum _{x=k+1}^{k(n+1)}{\frac {1}{x}}}$ 

for ${\displaystyle \Delta x=n/k}$  and ${\displaystyle x_{i}}$  is a sample point in each interval.

Series representation

The natural logarithm ${\displaystyle \ln(1+x)}$  has a well-known Taylor series^[22] expansion that converges for ${\displaystyle x}$  in the open-closed interval ${\displaystyle (-1,1]}$ :

${\displaystyle \ln(1+x)=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}x^{n}}{n}}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}-{\frac {x^{6}}{6}}+\cdots .}$ 

Within this interval, for ${\displaystyle x=1}$ , the series is conditionally convergent, and for all other values, it is absolutely convergent. For ${\displaystyle x>1}$  or ${\displaystyle x\leq -1}$ , the series does not converge to ${\displaystyle \ln(1+x)}$ . In these cases, different representations^[23] or methods must be used to evaluate the logarithm.

Harmonic number difference

It is not uncommon in advanced mathematics, particularly in analytic number theory and asymptotic analysis, to encounter expressions involving differences or ratios of harmonic numbers at scaled indices^[24]. The identity involving the limiting difference between harmonic numbers at scaled indices and its relationship to the logarithmic function provides an intriguing example of how discrete sequences can asymptotically relate to continuous functions. This identity is expressed as^[25]

${\displaystyle \lim _{k\to \infty }(H_{k(n+1)}-H_{k})=\ln(n+1)}$ 

which characterizes the behavior of harmonic numbers as they grow large. This approximation (which precisely equals ${\displaystyle \ln(n+1)}$  in the limit) reflects how summation over increasing segments of the harmonic series exhibits integral properties, giving insight into the interplay between discrete and continuous analysis. It also illustrates how understanding the behavior of sums and series at large scales can lead to insightful conclusions about their properties. Here ${\displaystyle H_{k}}$  denotes the ${\displaystyle k}$ -th harmonic number, defined as

${\displaystyle H_{k}=\sum _{j=1}^{k}{\frac {1}{j}}}$ 

The harmonic numbers are a fundamental sequence in number theory and analysis, known for their logarithmic growth. This result leverages the fact that the sum of the inverses of integers (i.e., harmonic numbers) can be closely approximated by the natural logarithm function, plus a constant, especially when extended over large intervals^[26][24][27]. As ${\displaystyle k}$  tends towards infinity, the difference between the harmonic numbers ${\displaystyle H_{k(n+1)}}$  and ${\displaystyle H_{k}}$  converges to a non-zero value. This persistent non-zero difference, ${\displaystyle \ln(n+1)}$ , precludes the possibility of the harmonic series approaching a finite limit, thus providing a clear mathematical articulation of its divergence^[28][29]. The technique of approximating sums by integrals (specifically using the integral test or by direct integral approximation) is fundamental in deriving such results. This specific identity can be a consequence of these approximations, considering:

${\displaystyle \sum _{j=k+1}^{k(n+1)}{\frac {1}{j}}\approx \int _{k}^{k(n+1)}{\frac {dx}{x}}}$ 

Harmonic limit derivation

The limit explores the growth of the harmonic numbers when indices are multiplied by a scaling factor and then differenced. It specifically captures the sum from ${\displaystyle k+1}$  to ${\displaystyle k(n+1)}$ :

${\displaystyle H_{k(n+1)}-H_{k}=\sum _{j=k+1}^{k(n+1)}{\frac {1}{j}}}$ 

This can be estimated using the integral test for convergence, or more directly by comparing it to the integral of ${\displaystyle 1/x}$  from ${\displaystyle k}$  to ${\displaystyle k(n+1)}$ :

${\displaystyle \lim _{k\to \infty }\sum _{j=k+1}^{k(n+1)}{\frac {1}{j}}=\int _{k}^{k(n+1)}{\frac {dx}{x}}=\ln(k(n+1))-\ln(k)=\ln \left({\frac {k(n+1)}{k}}\right)=\ln(n+1)}$ 

As the window's lower bound begins at ${\displaystyle k+1}$  and the upper bound extends to ${\displaystyle k(n+1)}$ , both of which tend toward infinity as ${\displaystyle k\to \infty }$ , the summation window encompasses an increasingly vast portion of the smallest possible terms of the harmonic series (those with astronomically large denominators), creating a discrete sum that stretches towards infinity, which mirrors how continuous integrals accumulate value across an infinitesimally fine partitioning of the ___domain. In the limit, the interval is effectively from ${\displaystyle 1}$  to ${\displaystyle n+1}$  where the onset ${\displaystyle k}$  implies this minimally discrete region.

Double series formula

The harmonic number difference formula for ${\displaystyle \ln(m)}$  is an extension^[25] of the classic, alternating identity of ${\displaystyle \ln(2)}$ :

${\displaystyle \ln(2)=\lim _{k\to \infty }\sum _{n=1}^{k}\left({\frac {1}{2n-1}}-{\frac {1}{2n}}\right)}$ 

which can be generalized as the double series over the residues of ${\displaystyle m}$ :

${\displaystyle \ln(m)=\sum _{r\in \mathbb {Z} _{m}\cap \mathbb {N} }\sum _{x\in \langle m\rangle \cap \mathbb {N} }\left({\frac {1}{x-r}}-{\frac {1}{x}}\right)=\sum _{r\in \mathbb {Z} _{m}\cap \mathbb {N} }\sum _{x\in \langle m\rangle \cap \mathbb {N} }{\frac {r}{x(x-r)}}}$ 

where ${\displaystyle \langle m\rangle }$  is the principle ideal generated by ${\displaystyle m}$ . Subtracting ${\displaystyle \textstyle {\frac {1}{x}}}$  from each term ${\displaystyle \textstyle {\frac {1}{x-r}}}$  (i.e., balancing each term with the modulus) reduces the magnitude of each term's contribution, ensuring convergence by controlling the series' tendency toward divergence as ${\displaystyle m}$  increases. For example:

${\displaystyle \ln(4)=\lim _{k\to \infty }\sum _{n=1}^{k}\left({\frac {1}{4n-3}}-{\frac {1}{4n}}\right)+\left({\frac {1}{4n-2}}-{\frac {1}{4n}}\right)+\left({\frac {1}{4n-1}}-{\frac {1}{4n}}\right)}$ 

This method leverages the fine differences between closely related terms to stabilize the series. The sum over all residues ${\displaystyle r\in \mathbb {N} }$  ensures that adjustments are uniformly applied across all possible offsets within each block of ${\displaystyle m}$  terms. This uniform distribution of the "correction" across different intervals defined by ${\displaystyle x-r}$  functions similarly to telescoping over a very large sequence. It helps to flatten out the discrepancies that might otherwise lead to divergent behavior in a straightforward harmonic series. Note that the structure of the summands of this formula matches those of the interpolated harmonic number ${\displaystyle H_{x}}$  when both the ___domain and range are negated (i.e., ${\displaystyle -H_{-x}}$ ). However, the interpretation and roles of the variables differ.

Deveci's Proof

A fundamental feature of the proof is the accumulation of the subtrahends ${\displaystyle \textstyle {\frac {1}{x}}}$  into a unit fraction, that is, ${\displaystyle \textstyle {\frac {m}{x}}={\frac {1}{n}}}$  for ${\displaystyle m\mid x}$ , thus ${\displaystyle m=\omega +1}$  rather than ${\displaystyle m=|\mathbb {Z} _{m}\cap \mathbb {N} |}$ , where the extrema of ${\displaystyle \mathbb {Z} _{m}\cap \mathbb {N} }$  are ${\displaystyle [0,\omega ]}$  if ${\displaystyle \mathbb {N} =\mathbb {N} _{0}}$  and ${\displaystyle [1,\omega ]}$  otherwise, with the minimum of ${\displaystyle 0}$  being implicit in the latter case due to the structural requirements of the proof. Since the cardinality of ${\displaystyle \mathbb {Z} _{m}\cap \mathbb {N} }$  depends on the selection of one of two possible minima, the integral ${\displaystyle \textstyle \int {\frac {1}{t}}dt}$ , as a set-theoretic procedure, is a function of the maximum ${\displaystyle \omega }$  (which remains consistent across both interpretations) plus ${\displaystyle 1}$ , not the cardinality (which is ambiguous^[30][31] due to varying definitions of the minimum). Whereas the harmonic number difference computes the integral in a global sliding window, the double series, in parallel, computes the sum in a local sliding window—a shifting ${\displaystyle m}$ -tuple—over the harmonic series, advancing the window by ${\displaystyle m}$  positions to select the next ${\displaystyle m}$ -tuple, and offsetting each element of each tuple by ${\displaystyle \textstyle {\frac {1}{m}}}$  relative to the window's absolute position. The sum ${\displaystyle \textstyle \sum _{n=1}^{k}\sum {\frac {1}{x-r}}}$  corresponds to ${\displaystyle H_{km}}$  which scales ${\displaystyle H_{m}}$  without bound. The sum ${\displaystyle \textstyle \sum _{n=1}^{k}-{\frac {1}{n}}}$  corresponds to the prefix ${\displaystyle H_{k}}$  trimmed from the series to establish the window's moving lower bound ${\displaystyle k+1}$ , and ${\displaystyle \ln(m)}$  is the limit of the sliding window (the scaled, truncated^[32] series):

${\displaystyle \sum _{n=1}^{k}\sum _{r=1}^{\omega }\left({\frac {1}{mn-r}}-{\frac {1}{mn}}\right)=\sum _{n=1}^{k}\sum _{r=0}^{\omega }\left({\frac {1}{mn-r}}-{\frac {1}{mn}}\right)}$ 

${\displaystyle =\sum _{n=1}^{k}\left(-{\frac {1}{n}}+\sum _{r=0}^{\omega }{\frac {1}{mn-r}}\right)}$ 

${\displaystyle =-H_{k}+\sum _{n=1}^{k}\sum _{r=0}^{\omega }{\frac {1}{mn-r}}}$ 

${\displaystyle =-H_{k}+\sum _{n=1}^{k}\sum _{r=0}^{\omega }{\frac {1}{(n-1)m+m-r}}}$ 

${\displaystyle =-H_{k}+\sum _{n=1}^{k}\sum _{j=1}^{m}{\frac {1}{(n-1)m+j}}}$ 

${\displaystyle =-H_{k}+\sum _{n=1}^{k}\left(H_{nm}-H_{m(n-1)}\right)}$ 

${\displaystyle =-H_{k}+H_{mk}}$ 

${\displaystyle \lim _{k\to \infty }H_{km}-H_{k}=\sum _{x\in \langle m\rangle \cap \mathbb {N} }\sum _{r\in \mathbb {Z} _{m}\cap \mathbb {N} }\left({\frac {1}{x-r}}-{\frac {1}{x}}\right)=\ln(\omega +1)=\ln(m)}$ 

Asymptotic identities

Pronic numbers

As a consequence of the harmonic number difference, the natural logarithm is asymptotically approximated by a finite series difference^[25], representing a truncation of the integral at ${\displaystyle k=n}$ :

${\displaystyle H_{2T[n]}-H_{n}\sim \ln(n+1)}$ 

where ${\displaystyle T[n]}$  is the nth triangular number, and ${\displaystyle 2T[n]}$  is the sum of the first n even integers. Since the nth pronic number is asymptotically equivalent to the nth perfect square, it follows that:

${\displaystyle H_{n^{2}}-H_{n}\sim \ln(n+1)}$ 

Prime number theorem

The prime number theorem provides the following asymptotic equivalence:

${\displaystyle {\frac {n}{\pi (n)}}\sim \ln n}$ 

where ${\displaystyle \pi (n)}$  is the prime counting function. This relationship is equal to^[25]: 2:

${\displaystyle {\frac {n}{H(1,2,\ldots ,x_{n})}}\sim \ln n}$ 

where ${\displaystyle H(x_{1},x_{2},\ldots ,x_{n})}$  is the harmonic mean of ${\displaystyle x_{1},x_{2},\ldots ,x_{n}}$ . This is derived from the fact that the difference between the ${\displaystyle n}$ th harmonic number and ${\displaystyle \ln n}$  asymptotically approaches a small constant, resulting in ${\displaystyle H_{n^{2}}-H_{n}\sim H_{n}}$ . This behavior can also be derived from the properties of logarithms: ${\displaystyle \ln n}$  is half of ${\displaystyle \ln n^{2}}$ , and this "first half" is the natural log of the root of ${\displaystyle n^{2}}$ , which corresponds roughly to the first ${\displaystyle \textstyle {\frac {1}{n}}}$ th of the sum ${\displaystyle H_{n^{2}}}$ , or ${\displaystyle H_{n}}$ . The asymptotic equivalence of the first ${\displaystyle \textstyle {\frac {1}{n}}}$ th of ${\displaystyle H_{n^{2}}}$  to the latter ${\displaystyle \textstyle {\frac {n-1}{n}}}$ th of the series is expressed as follows:

${\displaystyle {\frac {H_{n}}{H_{n^{2}}}}\sim {\frac {\ln {\sqrt {n}}}{\ln n}}={\frac {1}{2}}}$ 

which generalizes to:

${\displaystyle {\frac {H_{n}}{H_{n^{k}}}}\sim {\frac {\ln {\sqrt[{k}]{n}}}{\ln n}}={\frac {1}{k}}}$ 

${\displaystyle kH_{n}\sim H_{n^{k}}}$ 

and:

${\displaystyle kH_{n}-H_{n}\sim (k-1)\ln(n+1)}$ 

${\displaystyle H_{n^{k}}-H_{n}\sim (k-1)\ln(n+1)}$ 

${\displaystyle kH_{n}-H_{n^{k}}\sim (k-1)\gamma }$ 

for fixed ${\displaystyle k}$ . The correspondence sets ${\displaystyle H_{n}}$  as a unit magnitude that partitions ${\displaystyle H_{n^{k}}}$  across powers, where each interval ${\displaystyle \textstyle {\frac {1}{n}}}$  to ${\displaystyle \textstyle {\frac {1}{n^{2}}}}$ , ${\displaystyle \textstyle {\frac {1}{n^{2}}}}$  to ${\displaystyle \textstyle {\frac {1}{n^{3}}}}$ , etc., corresponds to one ${\displaystyle H_{n}}$  unit, illustrating that ${\displaystyle H_{n^{k}}}$  forms a divergent series as ${\displaystyle k\to \infty }$ .

Real Arguments

These approximations extend to the real-valued ___domain through the interpolated harmonic number. For example, where ${\displaystyle x\in \mathbb {R} }$ :

${\displaystyle H_{x^{2}}-H_{x}\sim \ln x}$ 

Stirling numbers

The natural logarithm is asymptotically related to the harmonic numbers by the Stirling numbers^[33] and the Gregory coefficients^[34]. By representing ${\displaystyle H_{n}}$  in terms of Stirling numbers of the first kind, the harmonic number difference is alternatively expressed as follows, for fixed ${\displaystyle k}$ :

${\displaystyle {\frac {s(n^{k}+1,2)}{(n^{k})!}}-{\frac {s(n+1,2)}{n!}}\sim (k-1)\ln(n+1)}$ 

Pascal's triangle

Extensions

To arbitrary bases

Isaac Newton once observed that the first five rows of Pascal's Triangle, considered as strings, are the corresponding powers of eleven. He claimed without proof that subsequent rows also generate powers of eleven.^[35] In 1964, Dr. Robert L. Morton presented the more generalized argument that each row ${\displaystyle n}$  can be read as a radix ${\displaystyle a}$  numeral, where ${\displaystyle \lim _{n\to \infty }11_{a}^{n}}$  is the hypothetical terminal row, or limit, of the triangle, and the rows are its partial products.^[36] He proved the entries of row ${\displaystyle n}$ , when interpreted directly as a place-value numeral, correspond to the binomial expansion of ${\displaystyle (a+1)^{n}=11_{a}^{n}}$ . More rigorous proofs have since been developed.^[37][38] To better understand the principle behind this interpretation, here are some things to recall about binomials:

• A radix ${\displaystyle a}$  numeral in positional notation (e.g. ${\displaystyle 14641_{a}}$ ) is a univariate polynomial in the variable ${\displaystyle a}$ , where the degree of the variable of the ${\displaystyle i}$ th term (starting with ${\displaystyle i=0}$ ) is ${\displaystyle i}$ . For example, ${\displaystyle 14641_{a}=1\cdot a^{4}+4\cdot a^{3}+6\cdot a^{2}+4\cdot a^{1}+1\cdot a^{0}}$ .
• A row corresponds to the binomial expansion of ${\displaystyle (a+b)^{n}}$ . The variable ${\displaystyle b}$  can be eliminated from the expansion by setting ${\displaystyle b=1}$ . The expansion now typifies the expanded form of a radix ${\displaystyle a}$  numeral,^[39][40] as demonstrated above. Thus, when the entries of the row are concatenated and read in radix ${\displaystyle a}$  they form the numerical equivalent of ${\displaystyle (a+1)^{n}=11_{a}^{n}}$ . If ${\displaystyle c=a+1}$  for ${\displaystyle c<0}$ , then the theorem holds for ${\displaystyle a=\{c-1,-(c+1)\}\;\mathrm {mod} \;2c}$  with odd values of ${\displaystyle n}$  yielding negative row products.^[41][42][43]

By setting the row's radix (the variable ${\displaystyle a}$ ) equal to one and ten, row ${\displaystyle n}$  becomes the product ${\displaystyle 11_{1}^{n}=2^{n}}$  and ${\displaystyle 11_{10}^{n}=11^{n}}$ , respectively. To illustrate, consider ${\displaystyle a=n}$ , which yields the row product ${\displaystyle \textstyle n^{n}\left(1+{\frac {1}{n}}\right)^{n}=11_{n}^{n}}$ . The numeric representation of ${\displaystyle 11_{n}^{n}}$  is formed by concatenating the entries of row ${\displaystyle n}$ . The twelfth row denotes the product:

${\displaystyle 11_{12}^{12}=1:10:56:164:353:560:650:560:353:164:56:10:1_{12}=27433a9699701_{12}}$ 

with compound digits (delimited by ":") in radix twelve. The digits from ${\displaystyle k=n-1}$  through ${\displaystyle k=1}$  are compound because these row entries compute to values greater than or equal to twelve. To normalize^[44] the numeral, simply carry the first compound entry's prefix, that is, remove the prefix of the coefficient ${\displaystyle \textstyle {n \choose n-1}}$  from its leftmost digit up to, but excluding, its rightmost digit, and use radix-twelve arithmetic to sum the removed prefix with the entry on its immediate left, then repeat this process, proceeding leftward, until the leftmost entry is reached. In this particular example, the normalized string ends with ${\displaystyle 01}$  for all ${\displaystyle n}$ . The leftmost digit is ${\displaystyle 2}$  for ${\displaystyle n>2}$ , which is obtained by carrying the ${\displaystyle 1}$  of ${\displaystyle 10_{n}}$  at entry ${\displaystyle k=1}$ . It follows that the length of the normalized value of ${\displaystyle 11_{n}^{n}}$  is equal to the row length, ${\displaystyle n+1}$ . The integral part of ${\displaystyle 1.1_{n}^{n}}$  contains exactly one digit because ${\displaystyle n}$  (the number of places to the left the decimal has moved) is one less than the row length. Below is the normalized value of ${\displaystyle 1.1_{1234}^{1234}}$ . Compound digits remain in the value because they are radix ${\displaystyle 1234}$  residues represented in radix ten:

${\displaystyle 1.1_{1234}^{1234}=2.885:2:35:977:696:\overbrace {\ldots } ^{\text{1227 digits}}:0:1_{1234}=2.717181235\ldots _{10}}$ 

Other proposals for this edit

Note: add this citation for "to integers" section, for second approach to extension, borrowing from Hilton and Pedersen's^[45]

The Value of a Row subsection under Rows will be replaced with the following:

The ${\displaystyle n}$ th row reads as the numeral ${\displaystyle 11_{a}^{n}}$  for all ${\displaystyle a}$ . See Extension to arbitrary bases.

The comment to this edit (the "Edit Summary") will be:

Replaced bullet point on powers of 11 with a more robust description. A discussion of this edit can be found on the Talk page.

1. "e", Wolfram MathWorld: ex. 17, 18, and 19, archived from the original on 2023-03-15.
2. Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks/Cole Cengage Learning. p. 742.
3. Bartle, Robert G.; Sherbert, Donald R. (2011). Introduction to Real Analysis (4th ed.). John Wiley & Sons. pp. 229–231. ISBN 978-0-471-43331-6. Since 1/2 ≤ L(h) ≤ U(h) ≤ 1/2, we conclude that L(h) = U(h) = 1/2. Therefore h is Darboux integrable on I = [0, 1] and ∫(0 to 1) h = ∫(0 to 1) x dx = 1/2.
4. Larson, Ron; Hodgkins, Anne (2017). "Section 11.4: Area and the Fundamental Theorem of Calculus". College Algebra and Calculus: An Applied Approach (2nd ed.). Boston, MA: Cengage Learning. Please see exercise 17.
5. This superparticular ratio can be interpreted as the ratio between the sum of the first k even integers and the first k odd integers. This ratio is ${\displaystyle \textstyle {\frac {k+1}{k}}}$  (corresponding to the upper Darboux integral, ${\displaystyle \textstyle {\sqrt[{\infty }]{e}}}$ ) or ${\displaystyle \textstyle {\frac {k-1}{k}}}$  (corresponding to the lower Darboux integral, ${\displaystyle \textstyle {\sqrt[{-\infty }]{e}}}$ ), depending on whether the even numbers start from two or from zero, respectively.
6. Brothers, Harlan (2012). "Pascal's Triangle: The Hidden Stor-e". The Mathematical Gazette. 96: 145–148. doi:10.1017/S0025557200004204.
7. Brothers, Harlan (2012). "Math Bite: Finding e in Pascal's Triangle". Mathematics Magazine. 85 (1): 51. doi:10.4169/math.mag.85.1.51.
8. Wolf, Marek (2019). "6+infinity new expressions for the Euler-Mascheroni constant". arXiv:1904.09855 [math.NT]. The above sum is real and convergent when zeros ${\displaystyle \rho }$  and complex conjugate ${\displaystyle {\bar {\rho }}}$  are paired together and summed according to increasing absolute values of the imaginary parts of ${\displaystyle \rho }$ . See formula 11 on page 3. Note the typographical error in the numerator of Wolf's sum over zeros, which should be 2 rather than 1.
9. Boya, L.J. (2008). "Another relation between π, e, γ and ζ(n)". Revista de la Real Academia de Ciencias Exactas, Físicas y Naturales. Serie A. Matemáticas. 102: 199–202. doi:10.1007/BF03191819. γ/2 in (10) reflects the residual (finite part) of ζ(1)/2, of course. See formulas 1 and 10.
10. Sondow, Jonathan (2005). "Double Integrals for Euler's Constant and ${\displaystyle \textstyle {\frac {4}{\pi }}}$  and an Analog of Hadjicostas's Formula". The American Mathematical Monthly. 112 (1): 61–65. doi:10.2307/30037385. JSTOR 30037385. Retrieved 2024-04-27.
11. Chen, Chao-Ping (2018). "Ramanujan's formula for the harmonic number". Applied Mathematics and Computation. 317: 121–128. doi:10.1016/j.amc.2017.08.053. ISSN 0096-3003. Retrieved 2024-04-27.
12. Lodge, A. (1904). "An approximate expression for the value of 1 + 1/2 + 1/3 + ... + 1/r". Messenger of Mathematics. 30: 103–107.
13. Villarino, Mark B. (2007). "Ramanujan's Harmonic Number Expansion into Negative Powers of a Triangular Number". arXiv:0707.3950 [math.CA]. It would also be interesting to develop an expansion for n! into powers of m, a new Stirling expansion, as it were. See formula 1.8 on page 3.
14. Mortici, Cristinel (2010). "On the Stirling expansion into negative powers of a triangular number". Math. Commun. 15: 359–364.
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16. Bromwich, Thomas John I'Anson (2005) [1908]. An Introduction to the Theory of Infinite Series (PDF) (3rd ed.). United Kingdom: American Mathematical Society. p. 460. See exercise 18.
17. Whittaker, E.; Watson, G. (2021) [1902]. A Course of Modern Analysis (5th ed.). p. 271, 275. doi:10.1017/9781009004091. ISBN 9781316518939. See Examples 12.21 and 12.50 for exercises on the derivation of the integral form ${\displaystyle \textstyle \int _{-1}^{0}\ln \Gamma (z+1)\,dz}$  of the series ${\displaystyle \textstyle \sum _{k=1}^{n}{\frac {\zeta (k)}{110_{k}}}=\ln({\sqrt {2\pi }})}$ .
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22. Weisstein, Eric W. "Mercator Series". MathWorld--A Wolfram Web Resource. Retrieved 2024-04-24.
23. To extend the utility of the Mercator series beyond its conventional bounds one can calculate ${\displaystyle \textstyle \ln(1+x)}$  for ${\displaystyle \textstyle x=-{\frac {n}{n+1}}}$  and ${\displaystyle \textstyle n\geq 0}$  and then negate the result, ${\displaystyle \textstyle \ln \left({\frac {1}{n+1}}\right)}$ , to derive ${\displaystyle \textstyle \ln(n+1)}$ . For example, setting ${\displaystyle \textstyle x=-{\frac {1}{2}}}$  yields ${\displaystyle \textstyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{n2^{n}}}}$ .
24. Flajolet, Philippe; Sedgewick, Robert (2009). Analytic Combinatorics. Cambridge University Press. p. 389. ISBN 978-0521898065. See page 117, and VI.8 definition of shifted harmonic numbers on page 389
25. Deveci, Sinan (2022). "On a Double Series Representation of the Natural Logarithm, the Asymptotic Behavior of Hölder Means, and an Elementary Estimate for the Prime Counting Function". arXiv:2211.10751 [math.NT]. See Theorem 5.2. on pages 22 - 23
26. Graham, Ronald L.; Knuth, Donald E.; Patashnik, Oren (1994). Concrete Mathematics: A Foundation for Computer Science. Addison-Wesley. p. 429. ISBN 0-201-55802-5.
27. "Harmonic Number". Wolfram MathWorld. Retrieved 2024-04-24. See formula 13.
28. Kifowit, Steven J. (2019). More Proofs of Divergence of the Harmonic Series (PDF) (Report). Prairie State College. Retrieved 2024-04-24. See Proofs 23 and 24 for details on the relationship between harmonic numbers and logarithmic functions.
29. Bell, Jordan; Blåsjö, Viktor (2018). "Pietro Mengoli's 1650 Proof That the Harmonic Series Diverges". Mathematics Magazine. 91 (5): 341–347. doi:10.1080/0025570X.2018.1506656. hdl:1874/407528. JSTOR 48665556. Retrieved 2024-04-24.
30. Harremoës, Peter (2011). "Is Zero a Natural Number?". arXiv:1102.0418 [math.HO]. A synopsis on the nature of 0 which frames the choice of minimum as the dichotomy between ordinals and cardinals.
31. Barton, N. (2020). "Absence perception and the philosophy of zero". Synthese. 197 (9): 3823–3850. doi:10.1007/s11229-019-02220-x. PMC 7437648. PMID 32848285. See section 3.1
32. The ${\displaystyle k+1}$  shift is characteristic of the right Riemann sum employed to prevent the integral from degenerating into the harmonic series, thereby averting divergence. Here, ${\displaystyle \textstyle -{\frac {1}{n}}}$  functions analogously, serving to regulate the series. The successor operation ${\displaystyle m=\omega +1}$  signals the implicit inclusion of the modulus ${\displaystyle m}$  (the region omitted from ${\displaystyle \mathbb {N} _{1}}$ ). The importance of this, from an axiomatic perspective, becomes evident when the residues of ${\displaystyle m}$  are formulated as ${\displaystyle e^{\ln(\omega +1)}}$ , where ${\displaystyle \omega +1}$  is bootstrapped by ${\displaystyle \omega =0}$  to produce the residues of modulus ${\displaystyle m=\omega =\omega _{0}+1=1}$ . Consequently, ${\displaystyle \omega }$  represents a limiting value in this context.
33. Khristo N. Boyadzhiev (2022). "New series identities with Cauchy, Stirling, and harmonic numbers, and Laguerre polynomials". arXiv. pp. 2, 6. arXiv:1911.00186. Retrieved 2023-11-06.
34. Comtet, Louis (1974). Advanced Combinatorics. Kluwer.
35. Newton, Isaac (1736), "A Treatise of the Method of Fluxions and Infinite Series", The Mathematical Works of Isaac Newton: 1:31–33, But these in the alternate areas, which are given, I observed were the same with the figures of which the several ascending powers of the number 11 consist, viz. ${\displaystyle 11^{0}}$ , ${\displaystyle 11^{1}}$ , ${\displaystyle 11^{2}}$ , ${\displaystyle 11^{3}}$ , ${\displaystyle 11^{4}}$ , etc. that is, first 1; the second 1, 1; the third 1, 2, 1; the fourth 1, 3, 3, 1; the fifth 1, 4, 6, 4, 1, and so on.
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