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.
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
(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 $(3)$.