$L^2$ convergence of partial sum of a sequence of functions: $\sum_{k=1}^n\frac{g^k(x)}{k}$ Let $g:\mathbb{R}\to \mathbb{C}$ be an $L^2$ function such that $|g(x)|\leq\epsilon<1, $ for every $x\in\mathbb{R}$, and $g(x) = O\left(\frac{1}{x}\right)$. I want to know if $h_n(x):= \sum_{k=1}^n \frac{g^k(x)}{k}$ converges in the $L^2$ sense. That it converges point wise follows from the fact that $h_n(x)$ is a Cauchy sequence for each $x$.
 A: Hint. From the classic identity
$$
\sum_{k=1}^\infty \frac{u^k}{k}=-\log(1-u),\quad |u|<1,
$$ one may observe that there exist $C>0$, $M>1$ such that
$$
\left| h_n(x)+\log(1-g(x))\right| =\left| \sum_{k=n+1}^\infty \frac{g^k(x)}{k}\right| \leq
    \begin{cases}
\left|\log(1-\epsilon)\right|,  & \text{if $|x|\leq M$ } \\[2ex]
\frac{C}{\large x}, & \text{if $|x|\geq M$ }
\end{cases}
$$ the latter function being clearly in $L^2\left(\mathbb{R}\right)$.
Observe now that, for $n\geq1$,
$$
\begin{align}
\left| \sum_{k=n+1}^\infty \frac{g^k(x)}{k}\right|\leq  \sum_{k=n+1}^\infty \frac{\left|g(x)\right|^k}{k}=|g(x)|^n\sum_{k=1}^\infty \frac{\left|g(x)\right|^k}{k+n}
\leq |g(x)|^n \times \left|\log(1-|g(x)|)\right|
\leq |g(x)|^{n+1}
\end{align}
$$ wich gives, for a certain fixed $M\geq 1$,
$$
\begin{align}
\left|\left| h_n(x)+\log(1-g(x))\right|\right|_2&=\int_{\mathbb{R}}\:\left| \sum_{k=n+1}^\infty \frac{g^k(x)}{k}\right|^2dx
\\\\&\leq \int_{\mathbb{R}}\:|g(x)|^{2n+2}\:dx
\\\\&\leq \int_{|x|\leq M}\:|g(x)|^{2n+2}\:dx+\int_{|x|>M}\:|g(x)|^{2n+2}\:dx
\\\\&\leq |\epsilon|^{2n}\int_{|x|\leq M}\:dx+|\epsilon|^{2n}\int_{|x|>M}\left|\frac1{x^2}\right|\:dx
\end{align}
$$
then, recalling that $0<\epsilon<1$, one gets the announced conclusion by letting $n \to \infty$.
