Evaluation of $\prod\limits_{n=1}^\infty(ne^{-n}+1)$

Question

We know that:

$$\prod_{n=1}^\infty(e^{-n}+1)=\left(-1,\frac1e\right)_\infty$$ with the Q Pocchhammer symbol $(a,q)_k$, but what if we made it:

$$P=\prod_{n=1}^\infty(ne^{-n}+1)= 2.27396845235252995…?$$

which does not appear in oeis. Here are some other forms of the constant:

$$\prod_{n=1}^\infty(ne^{-n}+1)=\lim_{k\to\infty}\frac{\prod\limits_{n=1}^k (e^n+n)}{e^\frac{k(k+1)}2}=\exp\left(\sum_{n=1}^\infty\ln(ne^{-n}+1)\right)=\exp\left(\lim_{k\to \infty}-\frac{k(k+1)}2+\sum_{n=1}^k \ln(e^n+n) \right)$$

If it helps, multiply over the product’s argument’s inverse’s range using the $-1$st branch of Lambert W:

$$\prod_{n=1}^\infty(ne^{-n}+1) = \prod_{-\text W_{-1}(n-1)=1}^\infty n$$

Using an integral:

$$\sum_{n=1}^\infty\ln(ne^{-n}+1)=-\int_0^\infty \lfloor x\rfloor d(\ln(xe^{-x}+1))=\int_0^\infty \frac{x\lfloor x\rfloor}{e^x+x}-\frac{\lfloor x\rfloor}{e^x+x} dx=0.8215265…$$

Additionally, expanding $\ln(x+1)$ and switching sums gives the polylogarithm function:

$$\ln(P)=-\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{n^k e^{-nk}}k=\sum_{n=1}^\infty\frac{(-1)^{n+1}}n\operatorname{Li}_{-n}(e^{-n})$$

$\operatorname{Li}_{-n}(x),n\in\Bbb N$ is a rational function which can be used.

Finally,

OEIS A022629, $a_n$, is about such products and gives $\displaystyle P=\sum_{n=0}^\infty a_ne^{-n}$

Is there a closed form or a new alternate form of the $2.27396845235252995…$ constant in terms of other sums, integrals, special functions, named constants etc?

Answer 1

Time to share one of the most beautiful things I learned on MSE. An approximated value for $\prod_{n\geq 0}\left(1+e^{-n}\right)$ can be derived from Mellin's inversion formula, as Marko Riedel shows here:

$$ \sum_{n\geq 1}\log\left(1-\frac{1}{2^{nx}}\right)\approx -\frac{\pi^2}{6x\log 2}+\log\sqrt{2\pi}-\frac{\log\log 2}{2}-\frac{\log x}{2}+\frac{\log 2}{24} x \tag{1}$$ this approximation is obscenely good. By picking $x=1/\log 2$ $$ \sum_{n\geq 1}\log\left(1-\frac{1}{e^n}\right)\approx -\frac{\pi^2}{6}+\log\sqrt{2\pi}+\frac{1}{24}\tag{2} $$ and by picking $x=2/\log 2$ $$ \sum_{n\geq 1}\log\left(1-\frac{1}{e^{2n}}\right)\approx -\frac{\pi^2}{12}+\log\sqrt{2\pi}-\frac{\log 2}{2}+\frac{1}{12} \tag{3} $$ If we consider the difference between $(3)$ and $(2)$ we get $$ \sum_{n\geq 1}\log\left(1+\frac{1}{e^n}\right)\approx \frac{\pi^2}{12}-\frac{\log 2}{2}+\frac{1}{24}\tag{4} $$ $$ \sum_{n\geq 0}\log\left(1+\frac{1}{e^n}\right) \approx \frac{\pi^2}{12}+\frac{\log 2}{2}+\frac{1}{24}\tag{5} $$ $$ \boxed{\prod_{n\geq 0}\left(1+\frac{1}{e^n}\right)\approx \sqrt{2}\, \exp\left(\frac{\pi^2}{12}+\frac{1}{24}\right)}\tag{6} $$ where the absolute error is less than $9\cdot 10^{-9}$.
I highly doubt there is a nice closed form, but $(6)$ should be more than enough for practical purposes.

Marko Riedel's technique can be adapted to treat $\sum_{n\geq 1}\log\left(1-\frac{n}{2^{nx}}\right)$, too. I do not like to steal credits, so I will leave him the task to fill the missing details.

Answer 2

Not as much of an evaluation as a number-theoretic interpretation.

Here is a very general formula. Given two countable sets $A,B$, and sufficiently nice $x_{a,b}$, we can write $$\prod_{b\in B}\left(\sum_{a\in A}x_{a,b}\right)=\sum_{\pi\in P(A,B)}\prod_{(a,b)\in\pi}x_{a,b},\tag1$$ where $P(A,B)$ is the set of all collections of points $\left\{(a_b,b):b\in B\right\}$ such that $a_b\in A$ for each $b\in B$. Here, $(x_{a,b})_{a\in A, b\in B}$ is "sufficiently nice" when the sums on the LHS of $(1)$ converge absolutely.

This formula looks bizarre, but it actually encodes a lot of information about partitions.

Anyway, setting $A=\{0,1\}$, $B=\Bbb Z_{\ge1}$, and $x_{a,b}=(be^{-b})^a$, we have $$P=\prod_{b\ge1}(1+be^{-b})=\sum_{\pi\in X}\prod_{(a,b)\in \pi}(be^{-b})^a,$$ where $X=P(\{0,1\},\Bbb Z_{\ge1})$. We can then see that $$\begin{align} P&=\sum_{\pi\in X}\prod_{(a,b)\in \pi}(be^{-b})^a\\ &=\sum_{\pi\in X}\left(\prod_{(a,b)\in \pi}e^{-ab}\right)\left(\prod_{(a,b)\in \pi}b^a\right)\\ &=\sum_{\pi\in X}e^{-|\pi|_{1,1}}\prod_{(a,b)\in \pi}b^a, \end{align}$$ where $|\pi|_{i,j}=\sum_{(a,b)\in\pi}a^ib^j$.

Now we take a look at $X=P(\{0,1\},\Bbb Z_{\ge1})$. Notice that every element of $X$ is of the form $\{(a_1,1),(a_2,2),(a_3,3),...\}$, where each $a_k\in\{0,1\}$. In particular, the set $\pi_0=\{(0,1),(0,2),(0,3),...\}$ is in $X$. We notice that $|\pi_0|_{1,1}=\sum_{(a,b)\in\pi_0}ab=\sum_{b\ge1}a_b b=\sum_{b\ge1}0=0,$ and that $\prod_{(a,b)\in\pi_0}b^a=\prod_{b\ge1}b^0=1.$ So, setting $X'=X\setminus\{\pi_0\},$ $$P=1+\sum_{\pi\in X'}e^{-|\pi|_{1,1}}\prod_{(a,b)\in\pi}b^{a}.$$ Now, we notice that for every $k\in\Bbb Z_{\ge1},$ the set $\pi_k=\{(0,1),(0,2),...,(0,k-1),(1,k),(0,k+1),...\}$ is in $X'$, and that $|\pi_k|_{1,1}=k$. Thus, we can reindex our sum $$P=1+\sum_{k\ge1}e^{-k}\sum_{\,\,\,\pi\in X'\\ |\pi|_{1,1}=k}\prod_{(a,b)\in\pi}b^a.$$ Note that we need not worry about the convergence of the product $\prod_{(a,b)\in\pi}b^a$, because if $\{(a,b)\in\pi: a=0\}$ is finite then the sum $|\pi|_{1,1}=\sum_{b\ge1}a_b b$ diverges to $+\infty$ and accordingly $e^{-|\pi|_{1,1}}\prod_{(a,b)\in\pi}b^a=0$, thus every such $\pi$ does not contribute to the sum $P$.

Anyway, this is significant, because each sum $$|\pi|_{1,1}=\sum_{{a,b}\in\pi}ab=\sum_{b\ge1}a_b b= N$$ is a partition of the integer $N$ into distinct parts, as $a_b\in\{0,1\}$ for each $b$. Furthermore, for each partition $|\pi|_{1,1}=b_1+b_2+...+b_m=N$, the corresponding product $\prod_{(a,b)\in\pi}b^a=b_1\cdot b_2\cdots b_m$ is the product of the parts. That is, the coefficient of $e^{-N}$ in the sum $P$ is equal to the sum of the products of the parts for each partition of $N$ into distinct parts. This is in agreement with the OEIS description of sequence A022629, which you mentioned in your question.