Prove coeffiecents of $\frac{1}{H(z)}$ are in $\ell^1(\mathbb{N})$ Let $H(z)=\sum_{k=0}^{\infty}\frac{1}{k+1}z^k$ which is analytic at D={$z\in\mathbb{C} : |z|<1$} , So- 
$$H(z)=\frac{ln(1-z)}{-z}$$ Hence H has no zeros in D and $G(z)=\frac{1}{H(z)}$ is also anaytic. And is representable by power series in 0 , i.e. ,$G(z)=\sum_{k=0}^{\infty}\alpha_kz^k$ .
Prove that $(\alpha_k)_1^\infty\in \ell^1(\mathbb{N})$.
I tried a lot of things , here is the two major-
1) $$G(z)=\frac{1}{1-(\frac{ln(1-z)}{z}+1)} \\=\sum_{k=0}^{\infty}(\frac{ln(1-z)}{z}+1)^k=\\=\sum_{k=0}^{\infty}((-1)\sum_{n=1}^{\infty}\frac{z^n}{n+1})^k=...$$(Also tried here to continue with the Binomial, didn't work )
2) Put $\beta_n=\frac{1}{n+1}$Tried to use that $\alpha_0=1 so- \forall n: \alpha_n+\sum_{1}^{n-1}\beta_{n-k}\alpha_k+\beta_n=0$ ,But it didn't help me either. 
 A: I don't think this question is elementary. 
Here are two possible routes. 
Hint 1. One may observe that 
$$
\ln \left( \frac{\ln (1+t)}{t} \right)  =   \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \left( \sum_{k = 1 }^{n } \frac{{n \brack k}}{k+1}\right) \frac{t^{n}}{n!}, \quad 0<t<1, \tag1
$$ where $\displaystyle {n \brack k}$ denote the Stirling numbers of the first kind (see a proof here). Then by differentiating with respect to $t$, one may use that
$$
\frac{t}{\ln(1+t)}=1+t+t(1+t)\cdot \frac{d}{dt} \ln \left( \frac{\ln (1+t)}{t} \right),  \tag2
$$ and deduce, from identity $(1)$, a series expansion for the left hand side of $(2)$. To prove that $\left\{\alpha_n\right\} \in \ell^2(\mathbb{N})$ may be deduced from the estimation

$$
\frac{1}{n!} \sum_{k = 1 }^{n } \frac{{n \brack k}}{k+1} = \mathcal{O} \left(\frac{1}{\ln n}\right), \quad \text{as} \quad n \rightarrow \infty,\tag3
$$ 

(proved here).
Hint 2. One may observe that
$$
\begin{align}
\frac{t}{\ln(1+t)}&=\int_0^1(1+t)^xdx, \qquad \quad |t|<1,
\\\\&=\sum_{n=0}^\infty \left[\int_0^1\binom {x}{n}\:dx\right] t^n
\\\\&=\sum_{n=0}^\infty \left(\int_0^1 x(x+1)\cdots (x+n-1)\:dx\right) \frac{t^n}{n!}
\end{align}
$$ one may then express the latter integrand in terms of the Stirling numbers of the first kind, concluding with $(3)$.
