Can we find a closed form for $\sum _{i=0}^{\infty } ((-1)^{i}x^i\prod_{j=1}^{i}\frac{e^j}{e^j-1})$? 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,
$$\frac{(-1)^nf^{(n)}(0)}{n!} = \prod_{i=1}^{n}\frac{e}{e^i - 1}$$
 A: 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:
$$ \frac{1}{(xe;e)_\infty} $$
A: $$ f(x) = \sum_{n=0}^{\infty} a_nx^n $$
$$ a_0 = 1 $$
$$ a_n = a_{n-1}\cdot \frac{e}{1-e^n} $$
$$ \frac{a_n}{a_{n-1}} = \frac{e}{1-e^n} = b_n $$
$$ a_n = \prod_{i=1}^{n} b_i $$
$$ b_n = e\prod_{i=1}^{n} \frac{1}{1-w_{i,n}e} = e\prod_{j=1}^{n} \frac{1}{1-e^{\frac{2\pi i j}{n}}e} = e\prod_{j=1}^{n} c_{j,n} $$
$$ c_{j,n} = \frac{1}{1-e^{1+\frac{2\pi i j}{n}}} = c_{kj,kn} $$
$$ a_n = e^n [c(1,1)]^n [c(1,2)]^{⌊\frac{n}{2}⌋}[c(1,3)c(2,3)]^{⌊\frac{n}{3}⌋}[c(1,4)c(3,4)]^{⌊\frac{n}{4}⌋}[c(1,5)c(2,5)c(3,5)c(4,5)]^{⌊\frac{n}{5}⌋}... $$
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:
$$ a_n^{-1} = a_{n-1}^{-1} e^{-1} \prod_{d|n} p_d(e) $$
$$ a_{10}^{-1} = a_{9}^{-1} e^{-1} p_1(e)p_2(e)p_5(e)p_{10}(e)$$
$$ = a_{9}^{-1} e^{-1} (1-e)(1+e)(1+e+e^2+e^3+e^4)(1-e+e^2-e^3+e^4) $$
