1. Statement of the problem.

1.1. We denote by $q$ a number of the form

where $2,3,5,\ldots,p$ are primes and and by $Q(x)$ the number of such numbers which do not exceed $x$: and our problem is that of determining the order of $Q(x)$. We prove that that is to say that to every positive $\epsilon$ corresponds an $x_0=x_0(\epsilon)$, such that for $x>x_0$. The function $Q(x)$ is thus of higher order than any power of $\log x$, but of lower order than any power of $x$.

The interest of the problem is threefold. In the first place the result itself, and the method by which it is obtained, are curious and interesting in themselves. Secondly, the method of proof is one which, as we shew at the end of the paper, may be applied to a whole class of problems in the analytic theory of numbers: it enables us, for example, to find asymptotic formulæ for the number of partitions of $n$ into positive integers, or into different positive integers, or into primes. Finally , the class of numbers $q$ includes as a sub-class the ``highly composite'' numbers recently studied by Mr.Ramanujan in an elaborate memoir in these Proceedings² The problem of determining, with any precision, the number $H(x)$ of highly composite numbers not exceeding $x$ appears to be one of extreme difficulty. Mr Ramanujan has proved, by elementary methods, that the order of $H(x)$ is at any rate greater than that of $\log x$³: but it is still uncertain whether or not the order of $H(x)$ is greater than that of any power of $\log x$. In order to apply transcendental methods to this problem, it would be necessary to study the properties of the function $$ {\mathfrak H}(s)=\sum\frac{1}{h^s}\, , $$ where $h$ is a highly composite number, and we have not been able to make any progress in this direction. It is therefore very desirable to study the distribution of wider classes of numbers which include the highly composite numbers and possess some at any rate of their characteristic properties. The simplest and most natural such class is that of the numbers $q$; and here progress is comparatively easy, since the function

possesses a product expression analogous to Euler's product expression for $\zeta(s)$, viz. where $l_n=2\cdot 3\cdot 5\cdots p_n$ is the product of the first $n$ primes.

We have not been able to apply to this problem the methods, depending on the theory of functions of a complex variable, by which the Prime Number Theorem was proved. The function ${\mathfrak Q}(s)$ has the line $\sigma=0$⁴ as a line of essential singularities, and we are not able to obtain sufficiently accurate information concerning the nature of these singularities. But it is easy enough to determine the behaviour of ${\mathfrak Q}(s)$ as a function of the real variable $s$; and it proves sufficient for our purpose to determine an asymptotic formula for ${\mathfrak Q}(s)$ when $s\to 0$, and then to apply a ``Tauberian'' theorem similar to those proved by Messrs Hardy and Littlewood in a series of papers published in these Proceedings and elsewhere⁵.

This ``Tauberian'' theorem is in itself of considerable interest as being (so far as we are aware) the first such theorem which deals with functions or sequences tending to infinity more rapidly than any power of the variable.

2. Elementary results.

2.1. Let us consider, before proceeding further, what information concerning the order of $Q(x)$ can be obtained by purely elementary methods.

Let

where $\vartheta(x) $ is Tschebyshef's function $$ \vartheta(x)=\sum_{p\leq x} \log p. $$ The class of numbers $q$ is plainly identical with the class of numbers of the form where $$b_1\geq 0,\, b_2\geq 0, \ldots, b_n\geq 0.$$

Now every $b$ can be expressed in one and only one way in the form

where every $c$ is equal to zero or to unity. We have therefore say, where

Let $r$ denote, generally, a number of the form

where every $c$ is zero or unity : and $R(x)$ the number of such numbers which do not exceed $x$. If $q\leq x$, we have $$ r_0\leq x, \:\: r_1^2\leq x, \:\: r_2^4\leq x,\, \ldots $$ The number of possible values of $r_0$, in formula (2.14), cannot therefore exceed $R(x)$; the number of possible values of $r_1$ cannot exceed $R(x^{\frac{1}{2}})$; and so on. The total number of values of $q$ can therefore not exceed where $\varpi$ is the largest number such that Thus

2.2. We denote by $f$ and $g$ the largest numbers such that

It is known⁶ (and may be proved by elementary methods) that constants $A$ and $B$ exist, such that and We have therefore $$e^{Ap}f\leq x,$$ $$ f\log f=O(\log x), $$ and \begin{align*} & \sum\limits_{1}^{g}\vartheta(p_\nu)\leq\log x,\quad \sum\limits_{1}^{g}p_\nu=O(\log x),\\ & \sum\limits_{1}^{g}\nu\log\nu=O(\log x), g^2\log g =O(\log x), \end{align*} But it is easy to obtain an upper bound for $R(x)$ in terms of $f$ and $g$. The number of numbers $l_1,l_2,\ldots,$ not exceeding $x$, is not greater than $f$; the number of products not exceeding $x$, of pairs of such numbers, is a fortiori not greater than $\frac{1}{2}f(f-1)$; and so on. Thus $$ R(x)\leq f+\frac{f(f-1)}{2!}+\frac{f(f-1)(f-2)}{3!}+\cdots , $$ where the summation need be extended to $g$ terms only, since $$ l_1l_2\cdots l_gl_{g+1}>x . $$ A fortiori, we have $$ R(x)\leq 1+f+\frac{f^2}{2!}+\cdots +\frac{f^g}{g!}\lt (1+f)^g=e^{g\log(1+f)}. $$ Thus by (2.23) and (2.24). Finally, since $$\log\sqrt{x}\log\log\sqrt{x}\lt \frac{1}{2}\log x\log\log x, $$ it follows from (2.16) and (2.17) that

2.3. A lower bound for $Q(x)$ may be found as follows. If $g$ is defined as in 2.2, we have $$l_1l_2\cdots l_g\leq x \lt l_1l_2\cdots l_gl_{g+1}.$$ It follows from the analysis of 2.2 that $$ l_{g+1}=e^{\vartheta(p_{g+1})}=e^{O(g\log g)} , $$ and $$\ds\qquad\qquad l_1l_2\cdots l_g=\exp\left\{\sum\limits_{1}^{g} \vartheta(p_\nu)\right\}=e^{O(g^2\log g)}. $$

Thus $$\ds x\lt e^{O(g^2\log g)};$$ which is only possible if $g$ is greater than a constant positive multiple of $$ \sqrt{\left(\frac{\log x}{\log\log x}\right)}. $$ Now the numbers $l_1,l_2,\ldots, l_g$ can be combined in $2^g$ different ways, and each such combination gives a number $q$ not greater than $x$. Thus

where $K$ is a positive constant. From (2.26) and (2.31) it follows that there are positive constants $K$ and $L$ such that The inequalities (2.32) give a fairly accurate idea as to the order of magnitude of $Q(x)$. But they are much less precise than the inequalities (1.121). To obtain these requires the use of less elementary methods.

3. The behaviour of $\mathfrak Q(s)$ when $s\to 0$ by positive values.

3.1. From the fact, already used in 2.1, that the class of numbers $q$ is identical with the class of numbers of the form (2.12), it follows at once that

Both series and product are absolutely convergent for $\sigma>0$, and where We have also

3.2. Lemma. If $x>1,\, s>0$, then

Write $x^s=e^u$: then we have to prove that

for all positive values of $u$ ; or (writing $w$ for $\frac{1}{2}u$) that for all positive values of $w$. But it is easy to prove that the function $$ f(w)=\frac{1}{w^2}-\frac{1}{\sinh^2 w} $$ is a steadily decreasing function of $w$, and that its limit when $w\to 0$ is $\frac{1}{3}$; and this establishes the truth of the lemma.

3.3. We have therefore

where

From the second of these inequalities, and (2.221), it follows that \begin{align*} \phi_1(s) \lt \frac{1}{s}\int_2^\infty\frac{e^{-Asx}}{(\log x)^2}\frac{dx}{x} &= \frac{e^{-2As}}{s\log 2}-A\int_2^\infty\frac{e^{-Asx}}{\log x}\ dx \\ &= \frac{1}{s\log 2}-A\int_2^\infty\frac{e^{-Asx}}{\log x}dx +O(1); \end{align*} and so that

On the other hand there is a positive constant $B$, such that \begin{equation*} \vartheta(x)\lt Bx\quad (x\geq 2)\href{#p34_en7}{^7}. \end{equation*}

Thus \begin{align*} \phi_1(s)>\frac{1}{s}\int_2^\infty\left\{\frac{1}{(\log x)^2}- \frac{s^2}{12}\right\}e^{-Bsx}\frac{dx}{x} &= \frac{1}{s}\int_2^\infty\frac{e^{-Bsx}}{(\log x)^2}\frac{dx}{x}+O(1)\\ &=\frac{1}{s\log 2}-B\int_2^\infty\frac{e^{-Bsx}}{\log x}dx +O(1); \end{align*} and so

3.4. Lemma. If $H$ is any positive number, then $$ J(s)=H\int_2^\infty\frac{e^{-Hsx}}{\log x}dx \sim \frac{1}{s\log(1/s)}, $$ when $s\to 0$.

Given any positive number $\epsilon$, we can choose $\xi$ and $X$, so that $$ \int_0^\xi He^{-Hx}dx\lt \epsilon, \quad \int_X^\infty He^{-Hx}dx\lt \epsilon. $$ Now \begin{align*} s\log\left(\frac{1}{s}\right)J(s)&= {\int_{2s}^\infty} \frac{He^{-Hu}\log(1/s)}{\log u+\log(1/s)}du ={\int_{2s}^{\sqrt{s}}}+ \int_{\sqrt{s}}^\xi+\int_\xi^X+\int_X^\infty\\ &= j_1(s)+j_2(s)+j_3(s)+j_4(s), \end{align*} say. And we have \begin{align*} 0 & \lt j_1(s)\lt \frac{\log(1/s)}{\log 2}\int_{2s}^{\sqrt{s}}He^{-Hu}du=O\{\sqrt{s}\log(1/s)\}=o(1),\\ 0 & \lt j_2(s)\lt 2\int_0^\xi He^{-Hu}du\lt 2\epsilon,\\ &\quad j_3(s)=\int_\xi^X He^{-Hu}du+o(1),\\ 0 & \lt j_4(s)\lt \int_X^\infty He^{-Hu} du\lt \epsilon; \end{align*} and so \begin{align*} \left|1-s\log\left(\frac{1}{s}\right)J(s)\right| &= \left|\int_0^\infty He^{-Hu}du-j_1(s)-j_2(s)-j_3(s)-j_4(s)\right|\\ &\lt 5\epsilon+o(1)\lt 6\epsilon, \end{align*} for all sufficiently small values of $s$.

3.5. From (3.32), (3.33), and lemma just proved, it follows that

From this formula we can deduce an asymptotic formula for $\log{\mathfrak{Q}}(s)$. We choose $N$ so that and we write say.

In the first place

In the second place $$ \phi(ns)=O\left\{\frac{1}{ns\log(1/ns)}\right\}, $$ and $$\log(1/ns)>\frac{1}{2}\log(1/s),$$ if $N\lt n\lt 1/\sqrt{s}$. It follows that a constant $K$ exists such that

Thirdly, $\sqrt{s}\leq ns\leq1$ in $\Phi_3(s)$, and a constant $L$ exists such that $$\phi(ns)\lt \frac{L}{\sqrt{s}\log(1/s)}\, . $$ Thus for all sufficiently small values of $s$.

Finally, in $\Phi_4(s)$ we have $ns>1$, and a constant $M$ exists such that $$ \phi(ns)\lt M2^{-ns}. $$ Thus

From (3.53), (3.541)–(3.544), and (3.52) it follows that where $$|\rho|\lt \epsilon\, , \, |\rho'|\lt K\epsilon.$$ Thus or

4. A Tauberian theorem.

4.1. The passage from (3.57) to (1.12) depends upon a theorem of the ``Tauberian'' type.

Theorem A. Suppose that

1. $\lambda_1\geq 0\, , \quad \lambda_n>\lambda_{n-1}\, , \quad \lambda_n\to\infty\, ;$
2. $\lambda_n/\lambda_{n-1}\to 1\, ;$
3. $a_n\geq 0\, ;$
4. $ A>0\:\: ,a>0\, ;$
5. $\sum a_ne^{-\lambda_ns}$ it is convergent for $s> 0 $;
6. $f(s)=\sum a_ne^{-\lambda_ns}=\exp\left[\{1+o(1)\}As^{-\alpha} \left\{\log\frac{1}{s}\right\}^{-\beta}\right]\, ,$

when $s\to 0$. Then $$ A_n=a_1+a_2+\cdots+a_n=\exp[\{1+o(1)\}B\lambda_n^{\alpha/(1+\alpha)} (\log\lambda_n)^{-\beta/(1+\alpha)}], $$ where $$B=A^{1/(1+\alpha)}\alpha^{-\alpha/(1+\alpha)} (1+\alpha)^{1+[\beta/(1+\alpha)]},$$ when $ n\to\infty$.

We are given that

for every positive $\delta$ and all sufficiently small values of $s$; and we have to shew that for every positive $\epsilon$ and all sufficiently large values of $n$.

In the argument which follows we shall be dealing with three variables, $\delta, s$ and $n$ (or $m$), the two latter variables being connected by an equation or by inequalities, and with an auxiliary parameter $\zeta$. We shall use the letter $\eta$, without a suffix, to denote generally a function of $\delta, s $ and $n$ (or $m$)⁸, which is not the same in different formulæ, but in all cases tends to zero when $\delta$ and $s$ tend to zero and $n$ (or $m$) to infinity; so that, given any positive $\epsilon$, we have $$ 0\lt |\eta|\lt \epsilon, $$ for $$ 0\lt \delta\lt \delta_0\, ,\quad 0\lt s\lt s_0\, ,\quad n>n_0. $$

We shall use the symbol $\eta_\zeta$ to denote a function of $\zeta$ only which tends to zero with $\zeta$, so that $$ 0\lt |\eta_\zeta|\lt \epsilon, $$ if $\zeta$ is small enough. It is to be understood that the choice of a $\zeta$ to satisfy certain conditions is in all cases prior to that of $\delta,s $ and $n$ (or $m$). Finally, we use the letters $H, K,\ldots$ to denote positive numbers independent of these variables and of $\zeta$.

The second of the inequalities (4.12) is very easily proved. For

where We can choose a value of $s$, corresponding to every large value of $n$, such that From these inequalities we deduce, by an elementary process of approximation, We have therefore for every positive $\epsilon$ and all sufficiently large values of $n$⁹.

4.2. We have

where ${\cal A}(x)$ is the discontinuous function defined by \begin{equation*} {\cal A}(x)=A_n\quad (\lambda_n\leq x\lt \lambda_{n+1})\href{#p34_en10}{^{10}}, \end{equation*} so that, by (4.16), for every positive $\epsilon$ and all sufficiently large values of $x$. We have therefore for every positive $\delta$ and all sufficiently small values of $s$.

We define $\lambda_x$, a steadily increasing and continuous function of the continuous variable $x$, by the equation $$ \lambda_x=\lambda_n+(x-n)(\lambda_{n+1}-\lambda_n)\quad (n\leq x\leq n+1). $$ We can then choose $m$ so that

We shall now shew that the limits of the integral in (4.23) may be replaced by $(1-\zeta)\lambda_m$ and $(1+\zeta)\lambda_m$, where $\zeta$ is an arbitrary positive number less than unity.

We write

where $H$ is a constant, in any case greater than 1, and large enough to satisfy certain further conditions which will appear in a moment; and we proceed to shew that $J_1, J_2, J_4 $ and $J_5$ are negligible in comparison with the exponentials which occur in (4.23), and so in comparison with $J_3$.

4.3. The integrals $J_1$ and $J_5$ are easily disposed of. In the first place we have

by (4.22)¹¹. It will be found, by a straightforward calculation, that this expression is less than and is therefore certainly negligible if $H$ is sufficiently large.

Thus $J_1$ is negligible. To prove that $J_5$ is negligible we prove first that $$ sx > 4Bx^{\alpha/(1+\alpha)}(\log x)^{-\beta/(1+\alpha)}, $$ if $x>H\lambda_m$ and $H$ is large enough¹². It follows that \begin{align*} J_5 = s\int_{H\lambda_m}^{\infty}{\cal A}(x) e^{-sx} dx & \lt s\int\limits_{H\lambda_m}^{\infty}\exp\{(1+\delta)Bx^{\alpha/(1+\alpha)}(\log x)^{-\beta/(1+\alpha)}-sx\} dx \\ &\lt s\int_{0}^{\infty}e^{-\frac{1}{2}sx}dx=\frac{1}{2}\, , \end{align*} and is therefore negligible.

4.4. The integrals $J_2$ and $J_4$ may be discussed in practically the same way, and we may confine ourselves to the latter. We have

where The maximum of the function $\psi$ occurs for $x=x_0$, where From this equation, and (4.24), it plainly results that and that $x_0$ falls (when $\delta$ and $s$ are small enough) between $(1-\zeta)\lambda_m$ and $(1+\zeta)\lambda_m$.

Let us write $x=x_0+\xi $ in $J_4$. Then $$ \psi(x)=\psi(x_0)+\frac{1}{2}(1+\delta)B\xi^2\frac{d^2}{dx_1^2} \{x_1^{\alpha/(1+\alpha)}(\log x_1)^{-\beta/(1+\alpha)}\}, $$ where $x_0\lt x_1\lt x$ and a fortiori $$ (1-\zeta)\lambda_m\lt x_1\lt H\lambda_m. $$ It follows that

On the other hand, an easy calculation shews that Thus Since $\zeta$ is independent of $\delta$ and $s$, this inequality shews that $J_4$ is negligible; and a similar argument may be applied to $J_2$.

4.5 We may therefore replace the inequalities (4.23) by

Since ${\cal A}(x)$ is a steadily increasing function of $x$, it follows that or If we substitute for $s$, in terms of $\lambda_m$, in the right-hand sides of (4.531) and (4.532), we obtain expressions of the form $$ \exp\{(1+\eta)B\lambda_m^{\alpha/(1+\alpha)} (\log\lambda_m)^{-\beta/(1+\alpha)}\}. $$ On the other hand $$e^{\zeta s\lambda_m}-e^{-\zeta s\lambda_m}$$ is of the form $$\exp\{\eta_\zeta\lambda_m^{\alpha/(1+\alpha)} (\log\lambda_m)^{-\beta/(1+\alpha)}\}.$$ We have thus Now let $\nu$ be any number such that Since $\lambda_n/\lambda_{n-1}\to 1$, it is clear that all numbers $n$ from a certain point onwards will fall among the numbers $\nu$. It follows from (4.541) and (4.542) that and therefore that, given $\epsilon$ we can choose first $\zeta$ and then $n_0$ so that for $n\geq n_0$. This completes the proof of the theorem.

4.6. There is of course a corresponding ``Abelian'' theorem, which we content ourselves with enunciating. This theorem is naturally not limited by the restriction that the coefficients $a_n$ are positive.

Theorem B. Suppose that

1. $\lambda_1\geq 0\, ,\quad \lambda_n>\lambda_{n-1}\, , \quad \lambda_n\to\infty\, ; $
2. $\lambda_n/\lambda_{n-1}\to 1\, ;$
3. $A>0\:\: ,0\lt\alpha\lt1\, ;$
4. $A_n =\alpha_1+\alpha_2+\cdots+\alpha_n =\exp\left[\{1+o(1)\}A\lambda_n^\alpha (\log\lambda_n)^{-\beta}\right]\, ,$

when $n\to\infty$. Then the series $\sum a_ne^{-\lambda_ns}$ is convergent for $s>0$, and $$ f(s)=\sum a_ne^{-\lambda_ns}=\exp[\{1+o(1)\}Bs^{-\alpha/(1-\alpha)} \left(\log\frac{1}{s}\right)^{-\beta/(1-\alpha)}], $$ where $$B=A^{1/(1-\alpha)}\alpha^{\alpha/(1-\alpha)} (1-\alpha)^{1+[\beta/(1-\alpha)]},$$ when $s\to 0$.

The proof of this theorem, which is naturally easier than that of the correlative Tauberian theorem, should present no difficulty to anyone who has followed the analysis which precedes.

4.7. The simplest and most interesting cases of Theorems A and B are those in which $$ \lambda_n=n\, , \quad \beta=0\, . $$ It is then convenient to write $x$ for $e^{-s}$. We thus obtain

Theorem C. If $A>0,\, 0\lt \alpha\lt 1$, and $$ \log A_n =\log(a_1+a_2+\cdots + a_n)\sim An^\alpha\, , $$ then the series $\sum a_nx^n$ is convergent for $|x|\lt 1$, and $$ \log f(x)=\log(\sum a_nx^n)\sim B(1-x)^{-\alpha/(1-\alpha)}\, , $$ where $$ B=(1-\alpha)\alpha^{\alpha/(1-\alpha)} A^{1/(1-\alpha)},$$ when $x\to 1$ by real values.

If the coefficients are positive the converse inference is also correct. That is to say, if $$ A>0\, , \alpha >0\, , $$ and $$ \log f(x)\sim A(1-x)^{-\alpha}\, ,$$ then $$ \log A_n\sim Bn^{\alpha/(1+\alpha)},$$ where $$ B=(1+\alpha)\alpha^{-\alpha/(1+\alpha)}A^{1/(1+\alpha)}.$$

5. Application to our problem, and to other problems in the Theory of Numbers.

5.1. We proved in 3 that

In Theorem A take $$ \lambda_n=\log n\, , \quad A=\frac{\pi^2}{6}\, , \quad \alpha=1\, , \quad \beta=1. $$ Then all the conditions of the theorem are satisfied. And $A_n$ is $Q(n)$, the number of numbers $q$ not exceeding $n$. We have therefore where

5.2. The method which we have followed in solving this problem is one capable of many other interesting applications.

Suppose, for example, that $R_r(n)$ is the number of ways in which $n$ can be represented as the sum of any number of $r^{th}$ powers of positive integers¹³.

We shall prove that

In particular, if $P(n)=R_1(n)$ is the number of partitions of $n$, then

We need only sketch the proof, which is in principle similar to the main proof of this paper. We have $$ \sum\limits_{1}^{\infty}R_r(n)e^{-ns}=\Pi_{1}^{\infty} \left(\frac{1}{1-e^{-s\nu^r}}\right), $$ and so

It is obvious that $R_r(n)$ increases with $n$ and that all the coefficients in $f(s)$ are positive. Again,

where But when $s\to 0$; and we can deduce, by an argument similar to that of 3.5, that We now obtain (5.21) by an application of Theorem A, taking $$ \lambda_n=n\, , \quad \alpha=\frac{1}{r}\, ,\quad \beta=0\, , \quad A=\Gamma\left(\frac{1}{r}+1\right)\zeta\left(\frac{1}{r}+1\right). $$ In a similar manner we can shew that, if $S(n)$ is the number of partitions of $n$ into different positive integers, so that \begin{align*} \sum S(n)e^{-ns}&= (1+e^{-s})(1+e^{-2s})(1+e^{-3s})+\cdots\\ &= \frac{1}{(1-e^{-s})(1-e^{-3s})(1-e^{-5s})\cdots}\, , \end{align*} then that if $T_r(n)$ is the number of representations of $n$ as the sum of $r^{th}$ powers of primes, then and, in particular, that if $T(n)=T_1(n)$ is the number of partitions of $n$ into primes, then

Finally, we can shew that if $r$ and $s$ are positive integers, $a>0$, and $0\leq b\leq 1$, and

then where In particular, if $a=1, b=1$, and $r=s$, we have

[Added March 28$^{th}$, 1917. – Since this paper was written M.G.Valiron (``Sur la croissance du module maximum des séries entières,'' Bulletin de la Société mathématique de France, Vol.XLIV, 1916, pp. 45 – 64) has published a number of very interesting theorems concerning power-series which are more or less directly related to ours. M.Valiron considers power-series only, and his point of view is different from ours, in some respects more restricted and in others more general.

He proves in particular that the necessary and sufficient conditions that $$ \log M(r)\sim\frac{A}{(1-r)^\alpha}, $$ where $M(r)$ is the maximum modulus of $f(x)=\sum a_nx^n$ for $|x|=r$, are that $$ \log|a_n|\lt (1+\epsilon)(1+\alpha)A^{1/(1+\alpha)}\left(\frac{n} {\alpha}\right)^{\alpha/(1+\alpha)} $$ for $n>n_0(\epsilon)$, and $$ \log|a_n|>(1-\epsilon_p)(1+\alpha)A^{1/(1+\alpha)} \left(\frac{n}{\alpha}\right)^{\alpha/(1+\alpha)} $$ for $n=n_p \, (p=1,2,3,\ldots)$, where $n_{p+1}/n_p\to 1$ and $\epsilon_p\to 0$ as $p\to\infty$.

M.Valiron refers to previous, but less general or less precise, results given by Borel (Leçons sur les séries à termes posítifs, 1902, Ch. V) and by Wiman (``Über dem Zusammenhang zwischen dem Maximal-betrage einer analytischen Funktion und dem grössten Gliede der zugehörigen Taylor'schen Reihe,'' Acta Mathematica, Vol.XXXVII, 1914, pp. 305 – 326). We may add a reference to Le Roy, ``Valeurs asymptotiques de certaines séries procédant suivant les puissances entières et positives d'une variable réelle,'' Bulletin des sciences mathématiques, Ser.2, Vol. XXIV, 1900, pp.245 – 268.

We have more recently obtained results concerning $P(n)$, the number of partitions of $n$, far more precise than (5.22). A preliminary account of these researches has appeared, under the title ``Une formule asymptotique pour le nombre des partitions de $n$,'' in the Comptes Rendus of January 2$^{nd}$, 1917¹⁵; and a fuller account has been presented to the Society. See Records of Proceedings at Meetings, March 1$^{st}$, 1917¹⁶.]