Unconventional way, how to expand to Maclaurin series

Question

Let's have function $f$ defined by: $$f(x)=2\sum_{k=1}^{\infty}\frac{e^{kx}}{k^3}-x\sum_{k=1}^{\infty}\frac{e^{kx}}{k^2},\quad x\in(-2\pi,0\,\rangle$$ My question: Can somebody expand it into a correct Maclaurin series, but using an unconventional way? Conventional is e.g. using $n$-th derivative of $f(x)$ in zero.

Reedited: Let me explain the reason for my question. This will be like conventionaly use of expansion of $e^{kx}$, but using incorrect arguments(using zetas for divergent series). Nice thing is, that the final result looks correct! We have: \begin{align}f(x)&=2\sum_{k=1}^{\infty}\sum_{m=0}^{\infty}\frac{k^{m-3}x^m}{m!}-\sum_{k=1}^{\infty}\sum_{m=0}^{\infty}\frac{k^{m-2}x^{m+1}}{m!}=\\&= \sum_{m=0}^{\infty}\frac{x^m}{m!}2\sum_{k=1}^{\infty}k^{m-3}-\sum_{m=0}^{\infty}\frac{x^{m+1}}{m!}\sum_{k=1}^{\infty}k^{m-2}=\\&=\sum_{m=0}^{\infty}\frac{x^m}{m!}2\zeta(3-m)-\sum_{m=0}^{\infty}\frac{x^{m+1}}{m!}\zeta(2-m)=\\&=\sum_{m=0}^{\infty}\frac{x^m}{m!}2\zeta(3-m)-\sum_{m=1}^{\infty}\frac{x^{m}}{(m-1)!}\zeta(3-m)=\\&=\sum_{m=0}^{\infty}\frac{x^m}{m!}(2-m)\zeta(3-m)\end{align} So we get nice Maclaurin series containing Zetas: $$f(x)=\sum_{m=0}^{\infty}\frac{x^m}{m!}(2-m)\zeta(3-m)$$ And now, if somebody will find expansion, but using some unconventional technique, there is a chance to get some interesting formula for $\zeta(3)$. That's motivation for my question.

Answer 1

let $g(x)=\sum_{k=1}^{\infty}\frac{e^{kx}}{k^3}$ then your problem is converted to $$f(x)=2g(x)-xg'(x)$$ then \begin{align*} f'(x)&=g'(x)-xg''(x)\\ f''(x)&=-x\cdot g'''(x) \end{align*} where $$g'''(x)=\sum_{k=1}^{\infty} e^{kx}$$ Thus $$f''(x)=-x\cdot\sum_{k=1}^{\infty} e^{kx}=-x\cdot\sum_{k=1}^{\infty}\sum_{m=0}^{\infty}\frac{(kx)^m}{m!}=-\sum_{k=1}^{\infty}\sum_{m=0}^{\infty}\frac{k^mx^{m+1}}{m!}$$ Now you can integrate. $$f(x)=-\sum_{k=1}^{\infty}\sum_{m=0}^{\infty}\frac{k^mx^{m+3}}{(m+3)(m+2)m!}$$

Answer 2

My first reaction and idea was to find a closed form for f(x) which I could later expand as a Taylor series. From definition, the function write
f(x) = -x PolyLog[2, E^x] + 2 PolyLog[3, E^x]
which is (it looks simple). However, the difficulties start when I try to expand this result as a Taylor series; the result is quite unpleasant and not really workable.

Meanwhile I was typing these first comments appeared Farshad Nahangi answer which is very nice and to which nothing has to be added if I want to avoid some stupid redundency.