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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.

Yarden Sharabi
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    Per Olivier's second hint, $\alpha_k < 0$ for all $k \geqslant 1$. So the series $\sum_{k = 0}^{\infty} \alpha_k$ converges either to a real number, or to $-\infty$. Looking sharply at (the proof of) Abel's theorem shows it's a real number (namely $0$). – Daniel Fischer May 22 '17 at 12:49
  • You are right, please help me prove that $\alpha_k < 0$ for all k⩾1 – Yarden Sharabi May 25 '17 at 22:37
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    If in Olivier's second hint you replace $t$ by $-z$, you'll see that $$\alpha_k = (-1)^k \int_0^1 \binom{x}{k},dx.$$ Then look at the sign of $\binom{x}{k}$ for $0 < x < 1$. (If you don't know what $\binom{x}{k}$ means when $x$ is not a nonnegative integer, reading this should help.) – Daniel Fischer May 26 '17 at 11:23

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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 $(3)$.

Olivier Oloa
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  • First, I realized that I needed something stronger, $(\alpha_k)_1^\infty\in \ell^1(\mathbb{N})$. – Yarden Sharabi May 22 '17 at 13:06
  • Second, I loved the second route, I can't finish it that way, maybe that way I can see that it doesn't true that $(\alpha_k)_1^\infty\in \ell^1(\mathbb{N})$? – Yarden Sharabi May 22 '17 at 13:09