On the series expansion of $\frac{\operatorname{Li}_3(-x)}{1+x}$ and its usage Recently I have asked about the evaluation of an integral involving a Trilogarithm $($which can be found here$)$. Pisco provided a quite an elegant approach starting with a functional equation of the Trilogarithm. However, his attempt made use of the fact that

$$\int_0^{\infty}\frac{\operatorname{Li}_3(-x)}{1+x}x^{s-1}\mathrm dx=\frac{\pi}{\sin \pi s}[\zeta(3)-\zeta(3,1-s)]~~~~~\text{for }0<s<1$$

Which can be seen as the Mellin Transform of the function $f(x)=\frac{\operatorname{Li}_3(-x)}{1+x}$. He also supplied a proof of his claim by using the Mellin Inversion Theorem. Anyway, I thought about applying Ramanujan's Master Theorem but for this purpose it is conclusive to  show that the given function, lets call it $f(x)$, has a power series of the form
$$f(x)=\sum_{n=0}^{\infty}(-1)^n\frac{\phi(n)}{n!}x^n$$
With the help of WolframAlpha I got that

$$f(x)=\frac{\operatorname{Li}_3(-x)}{1+x}=\frac12\sum_{n=1}^{\infty}[\psi^{(2)}(1+n)-\psi^{(2)}(1)](-x)^n~~\text{for }|x|<1$$

Using this result combined with Ramanujan's Master Theorem one can easily verify the given relation by setting $\phi(n)=\frac{\Gamma(1+n)}2[\psi^{(2)}(1+n)-\psi^{(2)}(1)]$ and applying the theorem
$$\begin{align}
\int_0^{\infty}\frac{\operatorname{Li}_3(-x)}{1+x}x^{s-1}\mathrm dx=\int_0^{\infty}f(x)x^{s-1}\mathrm dx&=\Gamma(s)\phi(-s)\\
&=\Gamma(s)\frac{\Gamma(1-s)}2[\psi^{(2)}(1-s)-\psi^{(2)}(1)]\\
&=\frac{\pi}{\sin \pi s}\frac12[(-2\zeta(3,1-s))-(-2\zeta(3))]\\
&=\frac{\pi}{\sin \pi s}[\zeta(3)-\zeta(3,1-s)]
\end{align}$$
Where Euler's Reflection Formula and some basic properties of the Polygamma Functions were used. This attempt affirms Pisco's proof. Up to now this is not a question but more a proof verification hence I am not sure whether all my steps are legitimate like this or not.

Since the power series given by WolframAlpha worked out fairly good I was curious how to derive such a series expansion. First of all I asked for a proof here to see how to approach to $f(x)$. Hence I was not able to identify a pattern within the coefficients of the MacLaurin expansion I thought about using a Cauchy Product here since both functions out of which $f(x)$ is composited have a well-known series expansion therefore this seems to be worth a try. Further the cited proof made use of it in some way as well. Thus, I tried to compute

$$\frac1{1+x}\cdot\operatorname{Li}_3(-x)=\left(\sum_{i=0}^{\infty}(-x)^i\right)\left(\sum_{j=1}^{\infty}\frac{(-x)^j}{j^3}\right)=~?$$

But I struggled to get started. I worried about the fact that the first index $i$ starts at $0$ while the second one $j$ at $1$. This would not be a big deal since I could just apply an index shift to one of the two series but then the powers of $(-x)$ would not longer match. Further, I am not totally sure how the author of referred proof dodged or accomplished this step.

My question is split up into two parts. $(1)$ Is my approach to the equality from above - beside the part where I "cheated" by asking WolframAlpha to expand $f(x)$ as a series - valid i.e. the usage of Ramanujan's Master Theorem here, the simplification of the RHS, etc.? $(2)$ How can one derive the series expansion of $f(x)$? Is it possible - without knowing the explicit power series - to deduce the pattern given by the MacLaurin expansion as values of the Polygamma Function  $\psi^{(2)}$? Can we apply the Cauchy Product here; if yes how can we take care of the unsuitable indices? I would be interested in a whole derivation of the series expansion for $f(x)$ as well.

Thanks in advance!
 A: Combining Maxim's suggestion to define the $0^{th}$ coefficient of the Trilogarithm series to be zero and ysharifi's given proof here on AoPS it is rather simple to obtain the given expansion. Using the Cauchy Product leads to
$$\begin{align}
\operatorname{Li}_3(-x)\cdot\frac1{1+x}&=\left(\sum_{n=1}^{\infty}\frac{(-x)^n}{n^3}\right)\cdot\left(\sum_{n=0}^{\infty}(-x)^n\right)\\
&=\sum_{n=1}^{\infty}\sum_{k=1}^n\frac1{k^3}(-x)^n\\
&=\sum_{n=1}^{\infty}\left[\sum_{k=1}^{\infty}\frac1{k^3}-\sum_{k=n+1}^{\infty}\frac1{k^3}\right](-x)^n\\
&=\sum_{n=1}^{\infty}\left[\sum_{k=0}^{\infty}\frac1{(1+k)^3}-\sum_{k=0}^{\infty}\frac1{(1+n+k)^3}\right](-x)^n\\
&=\sum_{n=1}^{\infty}\left[-\frac12\psi^{(2)}(1)+\frac12\psi^{(2)}(1+n)\right](-x)^n\\
\end{align}$$

$$\frac{\operatorname{Li}_3(-x)}{1+x}=\frac12\sum_{n=1}^{\infty}[\psi^{(2)}(1+n)-\psi^{(2)}(1)](-x)^n$$

A: Using the fact that $$\sum_{n=1}^\infty H_n^{(p)}x^n=\frac{\operatorname{Li}_p(x)}{1-x}$$
and by setting $p=3$ and replacing $x$ with $-x$, we get
$$\sum_{n=1}^\infty (-1)^nH_n^{(3)}x^n=\frac{\operatorname{Li}_3(-x)}{1+x}$$

Proof: 
Using $$\left(\sum_{n=1}^\infty a_nx^n\right)\left(\sum_{n=1}^\infty b_nx^n\right)=\sum_{n=1}^\infty x^{n+1}\left(\sum_{k=1}^na_k\ b_{n-k+1}\right)$$
Then
\begin{align}
\frac{\operatorname{Li}_p(x)}{1-x}&=\frac1{1-x}*\operatorname{Li}_p(x)\\
&=\sum_{n=1}^\infty x^{n-1}*\sum_{n=1}^\infty\frac{x^n}{n^p}\\
&=\frac1x\sum_{n=1}^\infty x^{n+1}\left(\sum_{k=1}^n\frac{1}{k^p}\right)\\
&=\sum_{n=1}^\infty H_n^{(p)}x^n
\end{align}
A: Different approach using a simple powerful identity
We proved here that
$$\sum_{n=1}^\infty a_nx^n=\frac1{1-x}\sum_{n=1}^\infty (a_n-a_{n-1})x^n,\quad a_{0}=0$$
Set $a_n=H_n^{(a)}$ we have 
$$\sum_{n=1}^\infty H_n^{(a)}x^n=\frac1{1-x}\sum_{n=1}^\infty (H_n^{(a)}-H_{n-1}^{(a)})x^n=\frac{1}{1-x}\sum_{n=1}^\infty \frac{x^n}{n^a}=\frac{\operatorname{Li}_a(x)}{1-x}$$
Set $a=3$ and replace $x$ with $-x$ to have 
$$\sum_{n=1}^\infty H_n^{(3)}(-x)^n=\frac{\operatorname{Li}_3(-x)}{1+x}$$
