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