It is connected with the logarithmic generating Function defined by: $$ \log(1+x)=\sum_{n \ge 1}(-1)^{n+1} \frac{x^n}{n} $$ The sequence is being generated by $a_0=0$ and $a_n=\frac{1}{n}$
Edit: I have used the Hint: from @Arnaud Mortier: $$ \frac{d}{dx} \sum_{n \ge 1} \frac{x^n}{n}=\sum_{n \ge 1} {x^{n-1}} $$
$$ \sum_{n \ge 1} x^{n-1}=\frac{1}{x} \sum_{n \ge 1} x^{n} $$
We know(if not p336 Concrete Mathematics)
$$ \sum_{n \ge 0} x^{n}=\frac{1}{1-x} $$ we also know(if not p335) that a multiplication with $x^1$ "shifts" the generating function to the right $$ \sum_{n \ge 1} x^{n}=x \frac{1}{1-x} $$ Now we have: $$ \sum_{n \ge 1} x^{n-1}=\frac{1}{1-x} $$ Now integrate both side( Now I have noticed i could have done right from the start since on page 335 there is this rule) $$ \int_{0}^{z} G(t) dt= \sum_{n \ge 1}\frac{1}{n}g_{n-1}z^n $$
$$ \sum_{n \ge 1} \frac{1}{n}x^{n-1}=\int_{0}^{x} \frac{1}{1-t}dt $$ Here I am not sure On the right side i get: $$ \int_{0}^{x} \frac{1}{1-t}dt=-ln(1-x) $$
I think some steps are obsolete, any mistakes? The integration used the substitution: u=1-x
Let $F(x) = \sum_{n=1}^\infty a_nx^n$ be the generating function.
Notice that your sequence satisfies $(n+1)a_{n+1} = na_n$ for all $n \ge 1$.
Multiplying this by $x^n$ and summing over $n \ge 1$ gives
$$F'(x) - a_1 = \sum_{n=1}^\infty (n+1)a_{n+1}x^n = \sum_{n=1}^\infty na_nx^n = xF'(x)$$
Rearranging gives $$F'(x) = \frac1{1-x} \implies F(x) = -\ln(1-x)$$