To find the ordinary power series generating function of $\left\{\frac{1}{n+1}\right\}_2^\infty$,
I tried to solve it like this, let $$\begin{align} f &= \frac{x^{n-2}}{n+1}, \text{ where }n \ge 2\\ f & = \frac{x^{n+1}}{n+1}\frac{1}{x^3} ,\\ f' & = \frac{x ^ {n}}{x^3} + \frac{x^{n+1}}{n+1} (-3)x^{-4}\\ f' & = x^{n-3} - \frac{ 3f}{x^2} , \end{align}$$ I don't know how to solve it further, usually the solution is obtained through integration however here if I try to integrate the $f$ term will be there still
The generating function for a sequence $\{ a_n \}$ is given by $f(x) = \sum_{n=0}^\infty a_n x^n$. In this case we have $$f(x) = \sum_{n=0}^\infty \frac{x^n}{(n+1)}.$$
We would like to find a way to express this in a closed form, and the geometric series will eventually come into play. Let's start by multiplying by $x$ and then taking a derivative:
$$xf(x) = \sum_{n=0}^\infty \frac{x^{n+1}}{n+1}$$ then $$(xf(x))' = f(x) + f'(x) x = \sum_{n=0}^\infty x^n = \frac{1}{1-x}.$$
Thus $$xf(x) = -\ln(1-x) + C.$$ Take $x=0$ and we see that $C=0$. Therefore $$f(x) = \frac{-\ln(1-x)}{x}$$
The reason we multiplied by $x$ was to match the power of $x$ in the series with the denominator. That way once we take the derivative, we will have simply a geometric series.
Finally, since the problem asked for the generating function corresponding to the terms starting at $n=2$, we can determine that from $f(x)$ by subtracting the first two terms: $$f(x) - 1 - \frac{x}{2}=\frac{-\ln(1-x)}{x} - 1 - \frac{x}2$$