1. Let the well-known expansion of $x \cot x$ (vide Edwards' Differential Calculus, $\S$149) be written in the form

from which we infer that $B_0$ may be supposed to be $-1$. Now \begin{eqnarray*} \cot x &= &\frac{\cos x}{\sin x} = \frac{\displaystyle 1 - \frac{x^2}{2!_{\vsp}} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots}{\displaystyle x - \frac{x^3{^{\vsp}}}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots }\\ &&\\ & =& \frac{\sin 2x}{1-\cos 2x} = \frac{\displaystyle\frac{2x}{1!_{\vsp}} - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \cdots}{\displaystyle\frac{(2x)^2}{2!} - \frac{(2x)^4}{4!} + \frac{(2x)^6}{6!} - \cdots }\\ &&\\ &=& \frac{1+ \cos 2x}{\sin 2x} = \frac{2 - \displaystyle\frac{(2x)^2}{2!_{\vsp}} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + \cdots} {\displaystyle 2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \frac{(2x)^7}{7!} + \cdots} \end{eqnarray*}

Multiplying both sides in each of the above three relations by the deno minator of the right-hand side and equating the coefficients of $x^n$ on both sides, we can write the results thus:

where $n$ is any odd integer; where $n$ is any even integer; where $n$ is any odd integer greater than unity.

From any one of (2), (3), (4) we can calculate the $B$'s. But as $n$ becomes greater and greater the calculation will get tedious. So we shall try to find simpler methods.

2. We know $$(x \cot x)^2 = -x^2 \left(1 + \frac{d \cot x}{dx} \right).$$ Using (1) and equating the coefficients of $x^n$ on both sides, and simplifying, we have $$ \half (n+1) B_n = c_2 B_2 B_{n-2} + c_4 B_4 B_{n-4} + c_6 B_6 B_{n-6} + \cdots,$$

A similar result can be obtained by equating the coefficients of $x^n$ in the identity $$\frac{d \tan x}{dx} = 1 + \tan^2 x.$$

3. Again \begin{eqnarray*} - \half x(\cot \half x + \coth \half x) &=& - \half x (\cot \half x+ i \cot \half ix) \\ &=& 2 \left\{B_0 + B_4 \frac{x^4}{4!} + B_8 \frac{x^8}{8!} + \cdots \right\}, \end{eqnarray*} by using (1). The expression may also be written \begin{eqnarray*} -\half x \frac{(\cos \half x \sin \half ix + i \sin \half x \cos \half ix)}{\sin \half x \sin \half ix} &=& - \half x \frac{(1+i) \sin \half x(1+i)-(1-i) \sin \half x (1-i)}{\cos \half x(1-i) - \cos \half x (1+i)}\\ &=& -x \frac{\strut \displaystyle\frac{x}{1!} - \frac{x^5}{2^2\cdot 5!} + \frac{x^9}{2^4\cdot 9!}-\cdots} {\displaystyle\frac{x^2}{2!} - \frac{x^6}{2^2\cdot 6!} + \frac{x^{10}}{2^4\cdot 10!}-\cdots} \end{eqnarray*} by expanding the numerator and the denominator, and simplifying by De Moivre's theorem.

Hence

Similarly

Proceeding as in $\S$1 we have, if $n$ is an even integer greater than 2, according as $n$ or $n-2$ is a multiple of 4. Analogous results can be obtained from $\tan \half x \pm \tanh \half x$. In (2), (3) and (4) there are $\half n$ terms, while in (5) and (8) there are $\frac{1}{4} n$ or $\frac{1}{4}(n-2)$ terms. Thus $B_n$ can be found from only half of the previous $B$'s.

4. A still simpler method can be deduced from the following identities.

If 1, $\omega, \omega^2$ be the three cube roots of unity, then $$4 \sin x \sin x \omega \sin x \omega^2 = - (\sin 2x + \sin 2 x \omega + \sin 2 x \omega^2),$$ as may easily be verified.

By logarithmic differentiation, we have $$\cot x + \omega \cot x \omega + \omega^2 \cot x \omega^2 = 2 \frac{\cos 2x + \omega \cos 2 x \omega + \omega^2 \cos 2 x \omega^2} {\sin 2 x + \sin 2 x \omega + \sin 2 x \omega^2}.$$ Writing $\half x$ for $x$, $$- \half x (\cot \half x + \omega \cot \half x \omega + \omega^2\cot \half x \omega^2) = - x\frac{\cos x + \omega \cos x \omega + \omega^2 \cos x \omega^2}{\sin x + \sin x \omega + \sin x \omega^2}$$ and, proceeding as in $\S$3, we get

Again $$\cot \half x \omega - \cot \half x \omega^2 = \frac{\cos x \omega^2 - \cos x \omega}{2 \sin \half x \sin \half x \omega \sin \half x \omega^2} = \frac{2(\cos x \omega - \cos x \omega^2)}{\sin x + \sin x \omega + \sin x \omega^2}.$$

Multiplying both sides by $- \half x(\omega^2 - \omega)$ and adding to the corresponding sides of the previous result, we have $$-\half x (\cot \half x + \omega^2 \cot \half x \omega + \omega \cot \half x \omega^2) = -x \frac{\cos x + \omega^2 \cos x \omega + \omega \cos x \omega^2} {\sin x + \sin x \omega + \sin x \omega^2}.$$ Hence, as before,

Similarly $$-x(\cot \half x + \cot \half x \omega + \cot \half x \omega^2) =x \frac{\cos x + \cos x \omega + \cos x \omega^2 - 3}{\sin x + \sin x \omega + \sin x \omega^2}, $$ and therefore

Multiplying up and equating coefficients in (9), (10) and (11) as usual, we have,

the last term being $\frac{1}{6}n(-1)^{\frac{1}{6}(n-1)}, \frac{1}{3}n(-1)^{\frac{1}{6}(n+1)}, \frac{1}{3}n(-1)^{\frac{1}{6}(n-3)}$.

Again, dividing both sides in (10) by $x$ and differentiating, we have \begin{eqnarray*} 3\left (B_2 \frac{1}{2!} + 7B_8 \frac{x^6}{8!} + 13 B_{14} \frac{x^{12}}{14!} +\cdots\right) &=& \frac{d}{dx} \left(\frac{\displaystyle\frac{x^4}{4!}-\frac{x^{10}}{10!}+ \frac{x^{16}}{16!}-\cdots}{\displaystyle\frac{x^3}{3!}-\frac{x^9}{9!} +\frac{x^{15}}{15!}-\cdots}\right)\\ &=&1-\frac{\displaystyle\frac{x^2}{2!}-\frac{x^8}{8!}+\frac{x^{14}}{14!}- \cdots}{\displaystyle\frac{x^3}{3!}-\frac{x^9}{9!}+\frac{x^{15}}{15!}-\cdots} \cdot\frac{\displaystyle\frac{x^4}{4!}-\frac{x^{10}}{10!}+\frac{x^{16}}{16!}- \cdots}{\displaystyle\frac{x^3}{3!}-\frac{x^9}{9!}+\frac{x^{15}}{15!}-\cdots}. \end{eqnarray*} Hence by (9) and (10), \begin{eqnarray*} &&3\left(B_2 \frac{x^2}{2!} + 7B_8 \frac{x^8}{8!} + 13 B_{14} \frac{x^{14}}{14!}+ \cdots\right)\\ &&=x^2+9\left(B_0+B_6 \frac{x^6}{6!}+B_{12}\frac{x^{12}}{12!}+\cdots\right) \left(B_2\frac{x^2}{2!} + B_8\frac{x^8}{8!} + B_{14}\frac{x^{14}}{14!}+\cdots\ \right). \end{eqnarray*}

Equating the coefficients of $x^n$ we have, if $n>2$ and $n-2$ is a multiple of 6,

From (12) the $B$'s can be calculated very quickly and (13) may prove useful in checking the calculations. The number of terms is one-third of that in (4); thus $B_{24}$ is found from $B_{18}, B_{12}$ and $B_6$.

5. We shall see later on how the $B$'s can be obtained from their properties only. But to know these properties, it will be convenient to calculate a few $B$'s by substituting $3,5,7,9,\cdots,$ for $n$ in succession in (12). Thus \begin{eqnarray*} &&B_0=-1; B_2 = \myfrac{1}{6}; B_4 = \myfrac{1}{30}; B_6= \myfrac{1}{42}; B_8 - \myfrac{1}{3} B_2 = - \myfrac{1}{45};\\ && \\ &&B_{10}- \myfrac{5}{2} B_4 = - \myfrac{1}{132}; B_{12} - 11 B_6 = - \myfrac{4}{455}; B_{14} -\myfrac{143}{4} B_8 + \frac{B_2}{5} = \myfrac{1}{120};\\ &&\\ &&B_{16}-\myfrac{286}{3} B_{10} + 4B_4 = \myfrac{1}{306}; B_{18}- 221 B_{12} + \myfrac{204}{5} B_6 = \myfrac{3}{665};\\ &&\\ &&B_{20}- \myfrac{3230}{7} B_{14} + \myfrac{1938}{7} B_8-\frac{B_2}{7}=-\myfrac{1}{231};\\ &&\\ &&B_{22}-\myfrac{3553}{4} B_{16} + \myfrac{7106}{5} B_{10} - \myfrac{11}{2} B_4=- \myfrac{1}{552}; \end{eqnarray*} and so on. Hence we have finally the following values: \begin{eqnarray*} &&B_2= \myfrac{1}{6}; B_4=\myfrac{1}{30}; B_6=\myfrac{1}{42}; B_8 =\myfrac{1}{30}; B_{10}= \myfrac{5}{66}; B_{12}=\myfrac{691}{2730}; B_{14}=\myfrac{7}{6};\\ &&\\ && B_{16}=\myfrac{3617}{510}; B_{18}=\myfrac{43867}{798}; B_{20}= \myfrac{174611}{330}; B_{22}=\myfrac{854513}{138}; B_{24}=\myfrac{236364091}{2730};\\ &&\\ &&B_{26}=\myfrac{8553103}{6}; B_{28}= \myfrac{23749461029}{870}; B_{30}=\myfrac{8615841276005}{14322};\\ &&\\ &&B_{32}=\myfrac{7709321041217}{510}; B_{34}=\myfrac{2577687858367}{6}; B_{36}=\myfrac{26315271553053477373}{1919190};\\ &&\\ &&B_{38}=\myfrac{2929993913841559}{6};~~ B_{40}=\myfrac{261082718496449122051}{13530}; \cdots, B_\infty=\infty. \end{eqnarray*}

6. It will be observed¹ that, if $n$ is even but not equal to zero,

\begin{equation} \mbox{(i) } B_n \mbox{ is a fraction and the numerator of } B_n/n \mbox{ in its lowest terms is a prime number, } \end{equation}

\begin{equation} \mbox{(ii) the denominator of } B_n \mbox{ contains each of the factors 2 and 3 once and only once,} \end{equation}

\begin{equation} \mbox{(iii) } 2^n(2^n-1)B_n/n \mbox{ is an integer and consequently } 2(2^n-1) B_n \mbox{ is an odd integer.} \end{equation}

\begin{equation} \begin{array}{l} \mbox{ From (16) it can easily be shown that the denominator of }\\ 2(2^n - 1) B_n/n \mbox{ in its lowest terms is the greatest power of 2 which divides }\\ n; \mbox{ and consequently, if } n \mbox{ is not a multiple of 4, then }\\ 4(2^n-1)B_n/n \mbox{ is an odd integer.} \end{array} \end{equation}

\begin{equation} \begin{array}{l} \mbox{ It follows from (14) that the numerator of }\\ B_n \mbox{ in its lowest terms is divisible by the greatest measure of }\\ n \mbox{ prime to the denominator, and the quotient is a prime number. } \end{array} \end{equation}

Examples: ($a$) 2 and 3 are the only prime factors of 12, 24 and 36 and they are found in the denominators of $B_{12}, B_{24}$ and $B_{36}$ and their numerators are prime numbers.

($b$) 11 is not found in the denominator of $B_{22}$, and hence its numerator is divisible by 11; similarly, the numerators of $B_{26}, B_{34}, B_{38}$ are divisible by 13, 17, 19, respectively and the quotients in all cases are prime numbers.

($c$) 5 is found in the denominator of $B_{20}$ and not in that of $B_{30}$, and consequently the numerator of $B_{30}$ is divisible by 5 while that of $B_{20}$ is a prime number. Thus we may say that if a prime number appearing in $n$ is not found in the denominator it will appear in the numerator, and vice versa.

7. Next, let us consider the denominators.

All the denominators are divisible by 6; those of $B_4, B_8, B_{12}, \cdots$ by 5; those of $B_6, B_{12},$ $B_{18}, \cdots$ by 7; those of $B_{10}, B_{20}, B_{30},\cdots$ by 11; but those of $B_8, B_{16}, B_{24}, \cdots $ are not divisible by 9; and those of $B_{14}, B_{28},\cdots$ are not divisible by 15. Hence we may infer that:

\begin{equation} \begin{array}{l} \mbox{ the denominator of $B_n$ is the continued product of prime numbers which are}\\ \mbox{ the next numbers (in the natural order) to the factors of } n\\ \mbox{ (including unity and the number itself)} \end{array} \end{equation}

As an example take the denominator of $B_{24}$. Write all the factors of 24, viz. 1, 2, 3, 4, 6, 8, 12, 24. The next numbers to these are 2, 3, 4, 5, 7, 9, 13, 25. Strike out the composite numbers and we have the prime numbers 2, 3, 5, 7, 13. And the denominator of $B_{24}$ is the product of 2, 3, 5, 7, 13, i.e., 2730.

It is unnecessary to write the odd factors of $n$ except unity, as the next numbers to these are even and hence composite.

The following are some further examples:

Even factors of $n$ and unity							Denominator of $B_n$
$B_2 $	$\cdt$	1, 2,	$\cdt$	$\cdt$	$\cdt$	$\cdt$	$2\cdot 3 = 6$
$B_6$	$\cdt$	1, 2, 6	$\cdt$	$\cdt$	$\cdt$	$\cdt$	$2\cdot 3\cdot 7=42$
$B_{12}$	$\cdt$	1, 2, 4, 6, 12		$\cdt$	$\cdt$	$\cdt$	$2\cdot 3\cdot 5\cdot 7\cdot 13=2730$
$B_{20}$	$\cdt$	1, 2, 4, 10, 20		$\cdt$	$\cdt$	$\cdt$	$2\cdot 3\cdot 5\cdot 11=330$
$B_{30}$	$\cdt$	1, 2, 6, 10, 30		$\cdt$	$\cdt$	$\cdt$	$2\cdot 3\cdot 7\cdot 11\cdot 31=14322$
$B_{42}$	$\cdt$	1, 2, 6, 14, 42		$\cdt$	$\cdt$	$\cdt$	$2\cdot 3\cdot 7\cdot 43=1806$
$B_{56}$	$\cdt$	1, 2, 4, 8, 14, 28, 56			$\cdt$	$\cdt$	$2\cdot 3\cdot 5\cdot 29=870$
$B_{72}$	$\cdt$	1, 2, 4, 6, 8, 12, 18, 24, 36, 72				$\cdt$	$2\cdot 3\cdot 5\cdot 7\cdot 13\cdot 19\cdot 37\cdot 73=140100870$
$B_{90}$	$\cdt$	1, 2, 6, 10, 18, 30, 90			$\cdt$	$\cdt$	$2\cdot 3\cdot 7\cdot 11\cdot 19\cdot 31=272118$
$B_{110}$	$\cdt$	1, 2, 10, 22, 110		$\cdt$	$\cdt$	$\cdt$	$2\cdot 3\cdot 11\cdot 23=1518$

8. Again taking the fractional part of any $B$ and splitting it into partial fractions, we see that:

\begin{equation} \begin{array}{l} \mbox{ the fractional part of } B_n=(-1)^{\frac{1}{2}n}\\ \{\mbox{the sum of the reciprocals of the prime factors of the denominator of } B_n\} - (-1)^{\frac{1}{2}n}. \end{array} \end{equation}

\begin{eqnarray*} \mbox{Thus the fractional part of } B_{16} &=& \half + \myfrac{1}{3} + \myfrac{1}{5} + \myfrac{1}{17} - 1 = \myfrac{47}{510};\\ \mbox{that of } B_{22} &=& 1 - \half -\myfrac{1}{3}-\myfrac{1}{23} =\myfrac{17}{138};\\ \mbox{that of } B_{28} &=& \half + \myfrac{1}{3} + \myfrac{1}{5} + \myfrac{1}{29} -1 = \myfrac{59}{870}; \mbox{ and so on.} \end{eqnarray*}

It can be inferred from (20) that:

\begin{equation} \begin{array}{l} \mbox{If } G \mbox{ be the G.C.M. and } L \mbox{ the L.C.M. of the denominators of}\\ B_m \mbox{ and } B_n, \mbox{ then } L/G \mbox{ is the denominator of } B_m-(-1)^{\frac{1}{2}(m-n)}B_n,\\ \mbox{ and hence, if the denominators of } B_m \mbox{ and } B_n \mbox{ are equal, then}\\ B_m-(-1)^{\frac{1}{2}(m-n)} B_n \mbox{ is an integer}. \end{array} \end{equation}

Example: $B_{24}-B_{12}$ and $B_{32}-B_{16}$ are integers, while the denominator of $B_{10} + B_{20}$ is 5.

It will be observed that:

\begin{equation} \begin{array}{l} \mbox{ (1) If } n \mbox{ is a multiple of 4, then the numerator of } B_n - \frac{1}{30}\\ \mbox{ in its lowest terms is divisible by 20; but if } n \mbox{ is not a multiple of 4 then}\\ \mbox{ that of } \frac{B_n}{n} - \frac{1}{12} \mbox{ in its lowest terms is divisible by 5; } \end{array} \end{equation}

(2) If $n$ is any integer, then

are integers of the form $30p+1$.

10. If a $B$ is known to lie between certain limits, then it is possible to find its exact value from the above properties.

Suppose we know that $B_{22}$ lies between 6084 and 6244; its exact value can be found as follows.

The fractional part of $B_{22} = \frac{17}{138}$ by (20), also $B_{22}$ is divisible by 11 by (18). And by (22) $B_{22} - \frac{11}{6}$ must be divisible by 5. To satisfy these conditions $B_{23}$ must be either $6137\frac{17}{138}$ or $6192\frac{17}{138}$.

But according to (18) the numerator of $B_{22}$ should be a prime number after it is divided by 11; and consequently $B_{22}$ must be equal to $6192\frac{17}{138}$ or $\frac{854513}{138}$, since the numerator of $6137\frac{17}{138}$ is divisible not only by 11 but also by 7 and 17.

11. It is known (Edward's Differential calculus, Ch. v, Ex.29) that $$B_n = \frac{2\cdot n!}{(2 \pi)^n} \left(\frac{1}{1^n}+ \frac{1}{2^n} + \frac{1}{3^n}+ \cdots \right),$$ or

where $2,3,5,\ldots$ are prime numbers.

Also

For \begin{eqnarray*} \int\limits^\infty_0 \frac{x^{n-1}}{e^{2 \pi x}-1} dx &=& \int\limits^\infty_0 x^{n-1} (e^{-2 \pi x} + e^{-4 \pi x} + \cdots) dx\\ &=& \frac{(n-1)!}{(2 \pi)^n} \left(\frac{1}{1^n} + \frac{1}{2^n}+\frac{1}{3^n} + \cdots \right)=\frac{B_n}{2n} \end{eqnarray*} by (24). In a similar manner $$\int\limits^\infty_0 \frac{x^n}{(e^{\pi x} - e^{-\pi x})^2} dx = \frac{B_n}{4 \pi},$$ and Take logarithms of both sides in (24) and write for $\log_e n!$ the well known expansion of $\log_e \Gamma(n+1),$ as in Carr's Synopsis, viz. where $0 < \theta < 1$, and where \begin{eqnarray*} \frac{B_{2p}\theta}{(2p-1)2pn^{2p-1}}&=& \frac{B_{2p}}{(2p-1)2pn^{2p-1}} -\frac{B_{2p+2}}{(2p+1)(2p+2)n^{2p+1}} + \cdots \\ &=&-\frac{1}{\pi}\int\limits^\infty_0 \frac{x^{2p-2}}{n^{2p-1}} \log(1-e^{-2 \pi x}) dx + \frac{1}{\pi} \int\limits^\infty_0 \frac{x^{2p}}{n^{2p+1}} \log (1-e^{-2 \pi x}) dx - \cdots\\ &=&-\frac{1}{\pi} \int\limits^\infty_0 \left(\frac{x^{2p-2}}{n^{2p-1}} -\frac{x^{2p}}{n^{2p+1}}+\cdots\right) \log(1-e^{-2 \pi x})dx\\ &=& - \frac{1}{\pi} \int\limits^\infty_0 \frac{x^{2p-2}}{n^{2p-3}(n^2+x^2)} \log (1-e^{-2 \pi x})dx \\ &=&-\int\limits^\infty_0 \frac{x^{2p-2}\log(1-e^{-2 \pi n x})}{\pi(1+x^2)}dx. \end{eqnarray*}

We can find the integral part of $B_n$, and since the fractional part can be found, as shewn in $\S$8, the exact value of $B_n$ is known. Unless the calculation is made to depend upon the values of $\log_{10}e, \log_e 10, \pi, \ldots$, which are known to a great number of decimal places, we should have to find the logarithms of certain numbers whose values are not found in the tables to as many places of decimals as we require. Such difficulties are removed by the method given in $\S$13.

12. Results (14) to (17), (20) and (21) can be obtained as follows. We have

\begin{eqnarray} \frac{1}{2x^2}&+&\frac{1}{(x+1)^2}+\frac{1}{(x+2)^2} + \frac{1}{(x+3)^2} +\frac{1}{(x+4)^2}+ \cdots\nonumber\\ &-&\frac{1}{x}-\frac{1}{6(x^3-x)}+\frac{1}{5(x^5-x)}+\frac{1}{7(x^7-x)} +\frac{1}{11(x^{11}-x)}+\cdots\nonumber\\ &=&\frac{1}{x^{15}}-\frac{7}{x^{17}}+\frac{55}{x^{19}} -\frac{529}{x^{21}}+\cdots \end{eqnarray}

where 5, 7, 11, 13, are prime numbers above 3. If we can prove that the left-hand side of (28) can be expanded in ascending powers of $1/x$ with integral coefficients, then (20) and (21) are at once deduced as follows.

From (27) we have

\begin{eqnarray} \frac{d^2 \log \Gamma(n+1)}{dn^2} &=& \frac{1}{(n+1)^2}+\frac{1}{(n+2)^2} +\frac{1}{(n+3)^2}+\cdots\nonumber\\ &=&\frac{1}{n}-\frac{1}{2n^2} + \frac{B_2}{n^3}-\frac{B_4}{n^5} +\frac{B_6}{n^7} -\frac{B_8}{n^9} + \cdots -(-)^p \frac{B_{2p}\theta}{n^{2p+1}}, \end{eqnarray} where \begin{eqnarray*} \frac{B_{2p}\theta}{n^{2p+1}}&=&\frac{B_{2p}}{n^{2p+1}} - \frac{B_{2p+2}}{n^{2p+3}} + \cdots\\ &=& 4 \pi \int\limits^\infty_0 \frac{x^{2p}}{n^{2p+1}(e^{\pi x}-e^{-\pi x})^2} dx - 4 \pi \int\limits^\infty_0 \frac{x^{2p+2}}{n^{2p+3}(e^{\pi x}-e^{-\pi x})^2} dx + \cdots\\ &=& \pi \int\limits^\infty_0 \left(\frac{x^{2p}}{n^{2p+1}}-\frac{x^{2p+2}}{n^{2p+3}} +\cdots \right) \frac{dx}{\sinh^2 \pi x}\\ &=&\pi \int\limits^\infty_0 \frac{x^{2p}}{n^{2p-1}(n^2+x^2)}\frac{dx}{\sinh^2 \pi x} =\int\limits^\infty_0 \frac{\pi x^{2p}}{(1+x^2)\sinh^2 \pi n x}dx. \end{eqnarray*} Substituting the result of (29) in (28) we see that $$\frac{B_2}{x^3}-\frac{B_4}{x^5}+\frac{B_6}{x^7}-\cdots- \frac{1}{6(x^3-x)}+\frac{1}{5(x^5-x)}+\frac{1}{7(x^7-x)}+\frac{1}{11(x^{11}-x)} +\cdots,$$ where $5,7,11,\ldots $ are prime numbers, can be expanded in ascending powers of $1/x$ with integral coefficients.

Therefore $B_2 - \frac{1}{6}$, $- B_4 - \frac{1}{6} + \frac{1}{5}$, $B_6 - \frac{1}{6} + \frac{1}{7}$, $-B_8 -\frac{1}{6} + \frac{1}{5}$, $B_{10}-\frac{1}{6} + \frac{1}{11}, \cdots,$ which are the coefficients of $1/x^3, 1/x^5, 1/x^7,\cdots,$ are integers.

Writing $\half + \frac{1}{3}-1$ for $-\frac{1}{6}$ we get the results of (20) and (21).

Again changing $n$ to $\half n$ in (29), and subtracting half of the result from (29), we have $$\frac{1}{(n+1)^2}-\frac{1}{(n+2)^2} + \frac{1}{(n+3)^2}-\cdots=\frac{1}{2n^2} -\frac{(2^2-1)B_2}{n^3}+\frac{(2^4-1)B_4}{n^5}$$

where $0 < \theta < 1,$ and also, by (29), $$ (2^{2p}-1) \frac{B_{2p}\theta}{n^{2p+1}} = \int\limits^\infty_0 \frac{\pi x^{2p} \cosh \pi n x}{(1+x^2) \sinh^2 \pi n x} dx.$$ Thus we see that, if we can prove that twice the left hand side of (30) can be expanded in ascending powers of $1/n$ with integral coefficients, then the second part of (16) is at once proved.

Again from (27) we have

\begin{eqnarray} \frac{d \log \Gamma (n+1)}{dn} & = & 1+\half +\myfrac{1}{3}+\myfrac{1}{4}+\cdots +\frac{1}{n} - \gamma \nonumber\\ & = &\log n + \frac{1}{2n}-\frac{B_2}{2n^2}+\frac{B_4}{4n^4}-\frac{B_6}{6n^6} +\frac{B_8}{8n^8}-\cdots + (-1)^p \frac{B_{2p}\theta}{2pn^{2p}}, \end{eqnarray}

where $0 < \theta < 1;$ and also, by (25), $$\frac{B_{2p}\theta}{2pn^{2p}} = \int\limits^\infty_0 \frac{2x^{2p-1}}{(1+x^2)(e^{2 \pi n x} - 1)} dx, $$ from which it can easily be shown that

\begin{eqnarray} \frac{1}{n+2} - \frac{1}{n+4} & + & \frac{1}{n+6}-\frac{1}{n+8} + \frac{1}{n+10} -\cdots \nonumber\\ &=&\frac{1}{2n} - 2(2^2-1) \frac{B_2}{2n^2} + 2^3 (2^4-1) \frac{B_4}{4n^4} - 2^5 (2^6-1) \frac{B_6}{6n^6}+ \cdots \nonumber\\ &&+(-1)^p 2^{2p-1} (2^{2p}-1) \frac{B_{2p}\theta}{2pn^{2p}}-\cdots, \end{eqnarray}

where $0 < \theta < 1;$ and also, by (31), $$2^{2p-1} (2^{2p}-1) \frac{B_{2p}\theta}{2pn^{2p}} = \int\limits^\infty_0 \frac{x^{2p-1}}{2(1+x^2)\sinh \half (\pi n x)} dx.$$ From the above theorem we see that, if we can prove that $$2 \left(\frac{1}{n+2} - \frac{1}{n+4} + \frac{1}{n+6}-\cdots\right),$$ can be expanded in ascending powers of $1/n$ with integral coefficients, then the first part of (16) at once follows.

13. The first few digits, and the number of digits in the integral part as well as in the numerator of $B_n$, can be found from the approximate formula: $$ \log_{10} B_n = (n+ \half ) \log_{10} n - 1.2324743503n + 0.700120, $$

This formula is proved as follows: taking logarithms of both sides in (24), $$ \log_e B_n = (n+\half ) \log_e n - n(1+\log_e 2 \pi) + \half \log_e 8 \pi $$ nearly. Multiplying both sides by $\log_{10} e$ or .4342944819, and reducing, we can get the result.

Changing $n$ to $n-2$ in (24) and taking the ratio of the two results, we have

where $2,3,5,\ldots$ are prime numbers.

\begin{equation} \begin{array}{l} \mbox{ Hence we see that } \frac{B_n}{B_{n-2}} \mbox{ approaches }\\ \frac{n(n-1)}{4 \pi^2} \mbox{ very rapidly as } n \mbox{ becomes greater and greater. } \end{array} \end{equation}

From the value of $\pi$, viz. 3.14159, 26535, 89793, 23846, 26433, 83279, 50288, 41971, 69399, $\ldots$, the integral part of any $B$ can be found from the previous $B$; and from the integral part the exact value can at once be written by help of (20) as follows:

Approximate ratio of any $B$ to previous $B$			lies between2				Hence the exact value is
$B_2$	$\cdt$	$\cdt$	0	and	1	$\cdt$	$1 -\frac{1^{\vsp}}{2} - \frac{1}{3}$	= $\frac{1}{6}$
$B_4$	$ = \frac{3\cdot 4}{4 \pi^2} B_2 $	$\cdt$	0	and	1	$\cdt$	$\half +\frac{1}{3} + \frac{1}{5}-1$	= $\frac{1}{30}$
$B_6$	$ =\frac{5\cdot 6}{4 \pi^2} B_4$	$\cdt$	0	and	1	$\cdt$	$1-\half -\frac{1}{3}-\frac{1}{7}$	= $\frac{1}{42}$
$B_8$	$=\frac{7\cdot 8}{4 \pi^2} B_6 $	$\cdt$	0	and	1	$\cdt$	$\half +\frac{1}{3}+\frac{1}{5}-1$	= $\frac{1}{30}$
$B_{10}$	$= \frac{9\cdot 10}{4 \pi^2} B_{8}$	$\cdt$	0	and	1	$\cdt$	$1- \half - \frac{1}{3}-\frac{1}{11}$	= $\frac{5}{66}$
$B_{12}$	$= \frac{11\cdot 12}{4 \pi^2} B_{10}$	$\cdt$	0	and	1	$\cdt$	$\half +\frac{1}{3}+\frac{1}{5} +\frac{1}{7}+\frac{1}{13}-1$	=$\frac{691}{2730}$
$B_{14}$	$= \frac{13\cdot 14}{4 \pi^2} B_{12}$	$\cdt$	0	and	2	$\cdt$	$2 - \half -\frac{1}{3} $	= $\frac{7}{6}$
$B_{16}$	$ =\frac{15\cdot 16}{4 \pi^2} B_{14}$	$\cdt$	7	and	8	$\cdt$	$6 + \half +\frac{1}{3} +\frac{1}{5}+\frac{1}{17}$	= $ \frac{3617}{510}$
$B_{18}$	$=\frac{17\cdot 18}{4 \pi^2} B_{16}$	$\cdt$	54	and	55	$\cdt$	$ 56-\half -\frac{1}{3}-\frac{1}{7}-\frac{1}{19}$	= $\frac{43867}{798}$
$B_{20}$	$ =\frac{19\cdot 20}{4 \pi^2} B_{18}$	$\cdt$	529	and	530	$\cdt$	$528+\half +\frac{1}{3}+\frac{1}{5}+\frac{1}{11}$	= $\frac{174611}{330}$
$\cdt$	$\cdt$	$\cdt$

From the preceding theorems we know some of the properties of $B_n$ for all positive even values of $n$. As an example let us take $B_{444}=N/D.$

The fractional part of $B_{444}$ is $\frac{23975417}{90709710}$ by (20). The numerator of $B_{444}$ is divisible by 37 and the quotient is a prime number by (18). Again $\log_{10} B_{444} = 630\cdot 2433$, nearly, by (33). Therefore the integral part of $B_{444}$ contains 631 digits, the first 4 digits being 1751. Again $$\log_{10} N = \log_{10} B_{444} + \log_{10} D=630\cdot 2433 + \log_{10} 90709710=638\cdot 2010 $$ nearly. Therefore $N$ contains 639 digits, the first four digits being 1588.

Again the numerator of $B_{444} - \frac{1}{30}$ is divisible by 20; that is to say, if $[B_{444}]$ is the integral part of $B_{444}$, $[B_{444}] +\frac{23975417}{90709710}-\frac{1}{30}=[B_{444}]+\frac{698392}{3023657}$ has a numerator divisible by 20. Therefore the integral part of $B_{444}$ ends with 4 and also the figure next to the last is even.

Hence $N$ ends with 57 and also the third figure from the last is even.

Instead of starting with $\cot x$ as in $\SS$ 2 and 3, we may start with $\tan x$ or $\mbox{ cosec } x$ and get other similar results.

Thus

\begin{eqnarray} (i)\:\:\myfrac{4}{3} B_n (2^n-1)& = & c_6 B_{n-6} (2^{n-6} - 1) - c_{12} B_{n-12} (2^{n-12}-1) + \cdots \nonumber\\ &+&\myfrac{1}{3} n (-1)^{\frac{1}{6}(n-2)} \:\:{\rm or}\:\: \myfrac{1}{6} n(-1)^{\frac{1}{6}(n-4)} \:\:{\rm or}\:\: \myfrac{1}{3}n(-1)^{\frac{1}{6}(n-6)} \cdots, \end{eqnarray}

\begin{eqnarray} (ii)c_3 \left(1-\frac{1}{2^{n-4}}\right) B_{n-3} - &&c_9 \left(1-\frac{1}{2^{n-10}}\right) B_{n-9} + c_{15} \left(1-\frac{1}{2^{n-16}}\right) B_{n-15} - \cdots\nonumber\\ & = & \ds{\myfrac{2}{3} \cdot \frac{n}{2^n} \{3^{\half (n-1)} + (-1)^{\frac{1}{6}(n-3)}\:\:{\rm or}\:\: (-1)^{\frac{1}{6}(n+1)}\:\:{\rm or}\:\: (-1)^{\frac{1}{6}(n+5)}\}}. \end{eqnarray}

The formulæ obtained in $\SS$1, 3, 4 may be called the one interval, two interval and three interval formula respectively. The $p$ interval formulæ can be got by taking the $p$th roots of unity or of $i$ according as $p$ is odd or even.

For example, let us take the fifth roots of unity $(1, \alpha, \alpha^2, \alpha^3, \alpha^4)$, and find the 5 interval formulæ.

Let \begin{eqnarray*} \phi (x) &=& \sin x + \sin x \alpha + \sin x \alpha^2 + \sin x \alpha^3 +\sin x \alpha^4 \\ &=& 5 \left(\frac{x^5}{5!} - \frac{x^{15}}{15!} + \frac{x^{25}}{25!}-\cdots \right). \end{eqnarray*} Then it can easily be shown that \begin{eqnarray*} 16 \sin x \sin x \alpha \sin x \alpha^2 \sin x \alpha^3 \sin x \alpha^4 &=& \phi(2x) - \phi \{2x (\alpha+\alpha^4)\} - \phi \{2x(\alpha^2+\alpha^3)\}\\ &=& \phi (2x) + \phi \{x(1 + \sqrt{5})\} + \phi \{x(1- \sqrt{5})\}. \end{eqnarray*} Taking logarithms and differentiating both sides, we have

where $$ \alpha_n = \left(\frac{1+\sqrt{5}}{2}\right)^n + \left(\frac{1-\sqrt{5}}{2}\right)^n, \mbox{ so that } \alpha_n \alpha_m = \alpha_{n+m} + (-1)^n \alpha_{m-n}. $$

Similarly

Again from $$16 \cos x \cos x \alpha \cos x \alpha^2 \cos x \alpha^3 \cos x \alpha^4 = 1 + \Psi [2x] + \Psi [2x (\alpha + \alpha^4)] + \Psi [2x (\alpha^2 + \alpha^3)]$$ where \begin{eqnarray*} \Psi(x) &=& \cos x + \cos x \alpha + \cos x \alpha^2 + \cos \alpha^3 + \cos x \alpha^4 \\ &=& 5 \left(1 - \frac{x^{10}}{10!} + \frac{x^{20}}{20!} - \frac{x^{30}}{30!} + \cdots \right), \end{eqnarray*} and similar relations, we can get many other identities.

18. The four interval formulæ can be got from the following identities: If $$ a_n= \left(1+ \frac{1}{\sqrt{2}}\right)^n + \left(1 - \frac{1}{\sqrt{2}}\right)^n \mbox{ and } b_n= \left(1+\frac{1}{\sqrt{2}}\right)^n - \left(1 - \frac{1}{\sqrt{2}}\right)^n, $$ so that $a_{m+n} = a_m a_n - a_{m-n}/ 2^n; \mbox{ and } b_{m+n} = a_m b_n + b_{m-n}/ 2^n$; then: