Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top



have any closed form in terms of known mathematical constants? The computer says $$p=3.682154\dots$$ but I don't even know how do devise the converging upper and lower bounds to obtain this result.

edit Jan. 15: I've got rid of the infinite product in favor of an fastly converging infinite sum over finite products here.


$$p=\lim_{n\to \infty}p_n\hspace{.7cm}\text{where}\hspace{.7cm} p_n=p_{n-1}\cdot \left(1+\frac{1}{n!}\right)\hspace{.7cm}\text{with}\hspace{.7cm} p_1=2.$$

So I looked for an emerging pattern



$p_3=((1+\frac{1}{1!}+\frac{1}{2!})+\frac{1}{1!2!})(1+\frac{1}{3!}) =(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!})+(\frac{1}{1!2!}+\frac{1}{1!3!}+\frac{1}{2!3!})+\frac{1}{1!2!3!}$

It appears that

$$p=1+\sum_{n=1}^\infty\sum_{m=1}^\infty a_{nm}$$

where $a_{1m}$ is the sum of terms with one inverse $\frac{1}{m!}$, and then $a_{2m}$ is the sum of (sums of) terms with two inverses $\frac{1}{r!s!}$. For example the term $\frac{1}{1!3!}$ is in the sum, and so I guess I need all the partitions into $n$ numbers. However, we don't want to count $\frac{1}{2!2!}$ and so it's more complicated. I guess the product can be written as a sum of term $(e-1)^n$ minus something, as for example

$(e-1)^2 = \left(\frac{1}{1!} + \frac{1}{2!}+ \frac{1}{3!}+\cdots\right)\left(\frac{1}{1!} + \frac{1}{2!}+ \frac{1}{3!}+\cdots\right) =\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{ m!\,n!}$.

The logarithm of it is also a sum of sums which somewhat resembles the series expansion of the exponential function, but there, I think, the coefficients are powers of $\frac{1}{n!}$.

share|cite|improve this question
It might be easier to compute $\prod (1+\frac{z^n}{n!})$, which should be $\sum \frac{1}{m!} P_m z^m$ where $P_m$ is the number of equivalence relations on a set of $m$ elements into classes of different sizes. – Thomas Andrews Aug 5 '13 at 20:11
I didn't mean "easier" in any pragmatic sense, just that an exponential generating function might be a useful approach to get some upper and lower bounds. As you can see, this function of $z$ counts something. It's not clear to me what use that might be. – Thomas Andrews Aug 5 '13 at 20:43
For a reference about the generating function check out page 137 of Phillip Flajolet's Analytic Combinatorics, entry II.26 – deoxygerbe Aug 5 '13 at 20:48
The ISC finds this number only in terms of various infinite products. – GEdgar Jan 13 '15 at 13:51 for $n\geq0$ – Fred Kline Jan 13 '15 at 13:54

Just one observation $$\ln p=\sum_{n=1}^{\infty}\ln \left(1+\frac{1}{n!}\right)<\sum_{n=1}^{\infty}\frac{1}{n!}=e-1$$ Since $$\ln (1+x)-x< 0$$ for all $x> 0$. So, $$p<e^{e-1}\approx 5.5749\ldots$$

Additional Observation: A tighter lower and upper bound comes as below:

$$\ln p=\sum_{n\ge 1}\ln\left(1+\frac{1}{n!}\right)=\ln 2+\sum_{n\ge 2}\sum_{k\ge 1}\frac{(-1)^{k-1}}{k(n!)^k}\\ $$ Then using the inequality $$\left(\sum_{i}a_i^k\right)\le \left(\sum_{i}a_i\right)^k$$ for $a_i\ge 0$, we get (after some calculations, which is not very difficult)$$\ln 2+e-2+\frac{1}{2}\ln (1-(e-2)^2)<\ln p<\ln 2+\frac{1}{2}\ln\left(\frac{e-1}{3-e}\right)\\ \Rightarrow 2e^{e-2}\sqrt{4e-e^2-3}<p<2\sqrt{\frac{e-1}{3-e}}\\ \Rightarrow 2.8538\ldots <p< 4.9393\ldots$$

share|cite|improve this answer

Note: Steven Stadnicki made a good point about the not-so-good convergence of my suggested computationa; method. I am modifying this to try to correct this.

Expanding Samrat's observation, for any positive integer $m$,

$\begin{align} \ln p &=\sum_{n=1}^{\infty}\ln \left(1+\frac{1}{n!}\right)\\ &=\sum_{n=1}^{m}\ln \left(1+\frac{1}{n!}\right)+\sum_{n=m+1}^{\infty}\ln \left(1+\frac{1}{n!}\right)\\ &= G_m+F_m\\ \text{where}\\ F_m &=\sum_{n=m+1}^{\infty}\ln \left(1+\frac{1}{n!}\right)\\ &=\sum_{n=m+1}^{\infty}\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k(n!)^k}\\ &=\sum_{k=1}^{\infty}\sum_{n=m+1}^{\infty} \frac{(-1)^{k-1}}{k(n!)^k}\\ &=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\sum_{n=m+1}^{\infty} \frac{1}{(n!)^k}\\ \end{align} $

Let $f_m(k) = \sum_{n=m+1}^{\infty} \frac{1}{(n!)^k} $, so $F_m = \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k} f_m(k) $.

$f_m(1)$ is $e$ minus the start of its series, and so is transcendental.

$f_m(2)$ is $I_0(2)$ minus the start of its series, where $I_0(x)$ is the modified Bessel function of the first kind. I do not know if $I_0(2)$ is transcendendal, but $J_0(1)$ is known to be, so I would be willing to bet that $all$ the $f_m(k)$ are transcendental.

I will now get an upper bound on $f_m(k)$ to aid in the computation of $F_m$.

$f_m(k) = \sum_{n=m+1}^{\infty} \frac{1}{(n!)^k} = \frac{1}{(m!)^k}\sum_{n=m+1}^{\infty} \left(\frac{m!}{n!}\right)^k $.

If $n > m$, $\frac{n!}{m!} =\prod_{k=1}^{n-m} (m+k) \ge (m+1)^{n-m} $, so $f_m(k) \le \frac{1}{(m!)^k}\sum_{n=m+1}^{\infty} \left(\frac{1}{(m+1)^{n-m}}\right)^k = \frac{1}{(m!)^k}\sum_{n=1}^{\infty} \frac{1}{(m+1)^{nk}} = \frac{1}{(m!)^k}\frac{(m+1)^{-k}}{1-(m+1)^{-k}} = \frac{1}{(m!)^k((m+1)^k-1)} $.

Since the $f_m(k)$ are decreasing, $|F_m| < f_m(1) < \frac{1}{m!m} $.

This means that the error in using the first $m$ terms in the product is less than this bound.

This somehow seems a more obvious result than I would like, but I will have to leave it at this since I do not see a more effective way of estimating the sum.

Possibly some acceleration method could be used to actually get more accurate estimates.

share|cite|improve this answer
The individual interior series converge quickly, but since they converge to 1 as $k\to\infty$, your 'exterior' series is still only converging as the $\frac{(-1)^k}{k}$ series for $\log 2$ does, i.e., terribly slowly (on the order of $n^{-1}$ for $n$ terms). – Steven Stadnicki Aug 5 '13 at 21:31
Good point. I will try to modify my answer to provide a better way of computing the result. – marty cohen Aug 6 '13 at 5:30

Using the general relation

$$a_1\cdot\prod_{n=1}^\infty\frac{a_{n+1}}{a_n}=a_1+\sum_{n=1}^\infty \left(\frac{a_{n+1}}{a_n}-1\right)a_n$$

with the partial products

$$a_n\equiv p_{n-1}:=\prod_{k=1}^{n-1}\left(1+\dfrac{1}{k!}\right),$$

for which it happens that


I found the sum representation


It is very fastly converging, as the monotonically increasing $a_n$ are, by definition, bounded by $p<4$. The fourth term is already $\dfrac{a_4}{4!}\approx 0.1\dots$

The sum can be written as


where $G$ is the Barnes G-function.

share|cite|improve this answer

I've been intrested in products aswell (see my question) the methode i used was this one, i hope it helps.

$$\prod_{i=b}^c 1+a_i$$= $$1+\sum_{i=b}^{c} (a_i)+$$ $$1/2!((\sum_{i=b}^c (a_i))^2-\sum_{i=b}^c (a_i)^2$$ $$1/3!((\sum_{i=b}^c (a_i))^3-3\sum_{i=b}^c (a_i)^2\sum_{i=b}^c (a_i)+2\sum_{i=b}^c (a_i)^3)$$ $$1/4!((\sum_{i=b}^c (a_i))^4-6(\sum_{i=b}^c (a_i))^2\sum_{i=b}^c (a_i)^2+3(\sum_{i=b}^c (a_i)^2)^2+8(\sum_{i=b}^c (a_i)^3)\sum_{i=b}^c (a_i)-6\sum_{i=b}^c (a_i)^4)$$

These are the refined strirling numbers, but i guess you get the pattern, i dont master the skills yet to write this more efficient down.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.