How can one prove that $$\sum_{n = 2}^{\infty} \left(\frac{(-1)^{n}}{(n-1)n} - \frac{9(-1)^n}{2\cdot 3^n}\right)x^n > 0\,?$$
this inequality resulted after expanding the difference of two functions into a power series, graphing both functions with a graphing clearly shows one above the other for all $x > 0$ so this means the series must be positive as well, but I can't see how one could prove it analytically.
Also it may be useful to note that first two terms of the power series are zero, so the series can start at $n=4$.
Let \begin{eqnarray} f(x)&=&\sum_{n = 2}^{\infty} \left(\dfrac{(-1)^{n}}{(n-1)n} - \dfrac{9(-1)^n}{2\cdot 3^n}\right)x^n\\ &=&\sum_{n = 2}^{\infty} \left(\dfrac{(-1)^{n}}{n-1}-\dfrac{(-1)^{n}}{n} - \dfrac{9(-1)^n}{2\cdot 3^n}\right)x^n\\ &=&-x+\ln(1+x)+x\ln(1+x)-\frac{3x^2}{2(3+x)}. \end{eqnarray} Now $$ f'(x)=\ln(1+x)-\frac{3x(6+x)}{2(3+x)^2}, f''(x)=\frac{x^2(9+x)}{2(1+x)(3+x)^2}. $$ Since $f''(x)>0$ for $x>0$, so $f'(x)>0$ for $x>0$ and hence $f(x)>0$ for $x>0$.