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Does

$$p=\prod_{n=1}^\infty\left(1+\frac{1}{n!}\right)$$

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.


Thoughts:

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

So I looked for an emerging pattern

$p_1=(1+\frac{1}{1!})$

$p_2=(1+\frac{1}{1!})(1+\frac{1}{2!})=(1+\frac{1}{1!}+\frac{1}{2!})+(\frac{1}{1!2!})$

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

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 $\frac{1}{n!}$.

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    $\begingroup$ 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. $\endgroup$ Commented Aug 5, 2013 at 20:11
  • $\begingroup$ 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. $\endgroup$ Commented Aug 5, 2013 at 20:43
  • 1
    $\begingroup$ For a reference about the generating function check out page 137 of Phillip Flajolet's algo.inria.fr/flajolet/Publications/books.html Analytic Combinatorics, entry II.26 $\endgroup$ Commented Aug 5, 2013 at 20:48
  • $\begingroup$ The ISC isc.carma.newcastle.edu.au finds this number only in terms of various infinite products. $\endgroup$
    – GEdgar
    Commented Jan 13, 2015 at 13:51
  • $\begingroup$ oeis.org/A238695 for $n\geq0$ $\endgroup$ Commented Jan 13, 2015 at 13:54

4 Answers 4

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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)$$\ln 2+e-2+\frac{1}{2}\ln (1-(e-2)^2)<\ln p<\ln 2+\frac{1}{2}\ln\left(\frac{e-1}{3-e}\right)\\ \Rightarrow 2e^{e-2}\sqrt{4e-e^2-3}<p<2\sqrt{\frac{e-1}{3-e}}\\ \Rightarrow 2.8538\ldots <p< 4.9393\ldots$$

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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.

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  • $\begingroup$ 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). $\endgroup$ Commented Aug 5, 2013 at 21:31
  • $\begingroup$ Good point. I will try to modify my answer to provide a better way of computing the result. $\endgroup$ Commented Aug 6, 2013 at 5:30
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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.

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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

$$\frac{a_{n+1}}{a_n}-1=\frac{1}{n!},$$

I found the sum representation

$$p=\sum_{n=0}^\infty\dfrac{1}{n!}\,{\prod_{k=1}^{n-1}}\left(1+\dfrac{1}{k!}\right)$$

The first few terms are $1,1,1,\tfrac{1}{2}$, for a total of 3.5. It is very fastly converging, as the monotonically increasing $a_n$ are, by definition, bounded by $p<4$. The fourth term is already $\dfrac{a_4}{4!}\approx 0.1\dots$

Here you can now also find a simple approximation scheme using the expansion of ${\rm e}^1$.

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