Find the coefficient of $z^k$ in $$\frac{\sum_{n\geq1} z^n/n}{1-z}$$
Is my solution correct? Write the above sum as:
$$\sum_{n\geq1} z^n/n\cdot\sum_{n\geq0} z^n$$
To get the coefficient of $z^k$ we simply see we can get k as a product of (coeff of z in first sum)(coeff of z in second sum): $(1,k-1)...(k,0)$. But the second sum has 1 as coefficient always.
So, Coefficient of $z^k$ in the whole sum is just : $$\sum_{n\geq1}^k1/n$$
Your answer is correct.
More generally if $a(x)=\sum_{n=1}^\infty a_nx^n$ and $b(x)=\sum_{n=0}^{\infty} b_nx^n$ then:
$$a(x)b(x)=\sum_{n=0}^\infty c_nx^n;\\c_n=\sum_{i=0}^n a_ib_{n-i}$$
The convergence in general of $\sum c_nx^n$ is only known for $x$ absolutely convergent in both $a(x)$ and $b(x),$ but it can be bigger than that, for example $a(x)=\frac1{1-x},b(x)=1-x.$
In your case, $a_0=0,$ so there are $n$ terms for $c_n$ rather than $n+1.$
This is called the Cauchy product of two series.
Your results, $\sum_{n=1}^k \frac1n,$ are often written as $H_k,$ and are called the harmonic numbers.