Can we find a closed form for $\sum _{i=0}^{\infty } ((-1)^{i}x^i\prod_{j=1}^{i}\frac{e^j}{e^j-1})$?

Question

Can we find a closed form for this infinite sum?

\begin{align*} f(x) &= \sum _{i=0}^{\infty } \biggl((-1)^{i}x^i\prod_{j=1}^{i}\frac{e}{e^j - 1} \biggr) \\ &= 1 - \biggl(\frac{e}{e - 1}\biggr)x + \biggl(\frac{e^2}{(e - 1)(e^2 - 1)} \biggr)x^2 \\ &\quad\hspace{6em} - \biggl(\frac{e^3}{(e - 1)(e^2 - 1)(e^3 - 1)} \biggr)x^3 + \dots \end{align*}

Each term is the sum of the sequence $\frac{1}{e^i}$ taken 1 by 1; 2 by 2; 3 by 3, and so on.

Or, also in other words:

$$f(x) = 1 -x \sum _{i_1=0}^{\infty } \frac{1}{e^{i_1}} +x^2 \sum _{i_1=0}^{\infty } \sum _{i_2=i_1 +1}^\infty \frac{1}{e^{i_1} e^{i_2}} - x^3 \sum _{i_1=0}^{\infty } \sum _{i_2=i_1+1}^\infty \sum _{i_3=i_2+1}^\infty \frac{1}{e^{i_1} e^{i_2} e^{i_3}} + ...$$

One other result, from Taylor series,

$$\frac{(-1)^nf^{(n)}(0)}{n!} = \prod_{i=1}^{n}\frac{e}{e^i - 1}$$

Answer 1

Q-Pochhammer symbol : Identities

$$ \frac{1}{(x;q)_\infty}=\sum_{n=0}^\infty \frac{x^n}{(q;q)_n} $$

$$ \frac{1}{(xe;e)_\infty}=\sum_{n=0}^\infty \frac{(xe)^n}{(e;e)_n} $$

$$ = \sum_{n=0}^\infty \frac{e^n}{(e;e)_n}x^n $$

$$ = \sum_{n=0}^\infty \frac{e^n}{(1-e)(1-e^2)...(1-e^n)}x^n $$

$$ = \sum_{n=0}^\infty (-1)^n\frac{e^n}{(e-1)(e^2-1)...(e^n-1)}x^n $$

$$ = \sum_{n=0}^\infty (-1)^nx^n\cdot\frac{e}{e-1} \cdot \frac{e}{e^2-1}...\frac{e}{e^n-1} $$

$$ = \sum_{n=0}^\infty (-1)^nx^n \prod_{j=1}^{n}\frac{e}{e^j-1} $$

Solution is:

$$ \frac{1}{(xe;e)_\infty} $$

Answer 2

$$ f(x) = \sum_{n=0}^{\infty} a_nx^n $$ $$ a_0 = 1 $$ $$ a_n = a_{n-1}\cdot \frac{e}{1-e^n} $$ $$ \frac{a_n}{a_{n-1}} = \frac{e}{1-e^n} = b_n $$ $$ a_n = \prod_{i=1}^{n} b_i $$ $$ b_n = e\prod_{i=1}^{n} \frac{1}{1-w_{i,n}e} = e\prod_{j=1}^{n} \frac{1}{1-e^{\frac{2\pi i j}{n}}e} = e\prod_{j=1}^{n} c_{j,n} $$ $$ c_{j,n} = \frac{1}{1-e^{1+\frac{2\pi i j}{n}}} = c_{kj,kn} $$ $$ a_n = e^n [c(1,1)]^n [c(1,2)]^{⌊\frac{n}{2}⌋}[c(1,3)c(2,3)]^{⌊\frac{n}{3}⌋}[c(1,4)c(3,4)]^{⌊\frac{n}{4}⌋}[c(1,5)c(2,5)c(3,5)c(4,5)]^{⌊\frac{n}{5}⌋}... $$ For example: $$ a_{13}^{-1} = e^{-13}(1-e)^{13}(1+e)^6(1+e+e^2)^4(1+e^2)^3(1+e+e^2+e^3+e^4)^2(p_6(e))^2...p_{12}(e)(1+e+e^2+e^3+...+e^{12})$$ $$ = e^{-13} p_1(e)^{13}p_2(e)^6p_3(e)^4p_4(e)^3p_5(e)^2p_6(e)^2p_7(e)...p_{13}(e)$$ where: $$ p_n(x) = \prod_{a}(1-u_n^ax); a<n; gcd(a,n)=1; u_n^a = e^{2\pi a i / n} $$ other formula: $$ a_n^{-1} = a_{n-1}^{-1} e^{-1} \prod_{d|n} p_d(e) $$ $$ a_{10}^{-1} = a_{9}^{-1} e^{-1} p_1(e)p_2(e)p_5(e)p_{10}(e)$$ $$ = a_{9}^{-1} e^{-1} (1-e)(1+e)(1+e+e^2+e^3+e^4)(1-e+e^2-e^3+e^4) $$