The series for harmonic numbers $H_k$ is:
$$ \frac{1}{1-z}\ln{\frac{1}{1-z}} $$
which then gives the series:
$$ 1z + \left(1+\frac{1}{2}\right)z^2 +\left(1+\frac{1}{2}+\frac{1}{3}\right)z^3 + \ldots $$
However, I encountered a problem which asks for the inverse of the harmonic series $1/H_k$ ... is there any series for the inverse of the harmonic numbers?
$$y=\frac1{1-z}\log \left(\frac{1}{1-z}\right)=\sum_{n=1}^\infty H_n\, z^n$$ Make the long division to get $$\frac{1}{z}-\frac{3}{2}+\frac{5 z}{12}+\frac{z^2}{24}+\frac{11 z^3}{720}+\frac{11 z^4}{1440}+\frac{271 z^5}{60480}+\frac{13 z^6}{4480}+\frac{7297 z^7}{3628800}+\frac{425 z^8}{290304}+O\left(z^9\right)$$
For large $n$, you can use $$\frac 1{H_n}=\frac{1}{\log (n)+\gamma }-\frac{1}{2 n (\log (n)+\gamma )^2}+\frac{\log (n)+\gamma +3}{12 n^2 (\log (n)+\gamma )^3}-$$ $$\frac{2 \log (n)+2 \gamma +3}{24 n^3 (\log (n)+\gamma )^4}+O\left(\frac{1}{n^4}\right)$$ For $n=10$ $$\frac 1{H_{10}}=\frac{2520}{7381}\approx 0.341417152$$ while the above would give $0.341417101$