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I am surprised by the question. The sums are bounded if $\alpha \ne 1$, since for every $n \ge 1$, $$\Big|\sum_{k=1}^n \Re R_\alpha^k z \Big| = \Big|\Re \Big(\sum_{k=1}^n R_\alpha^k z \Big) \Big| = \Big| \Re \Big(\frac{\alpha-\alpha^{n+1}}{1-\alpha}z\Big) \Big| \le \Big|\frac{\alpha-\alpha^{n+1}}{1-\alpha}z\Big| \le \frac{2}{|1-\alpha|}.$$ The log of this quantity is bounded above, so it can not increase like $(1/2)\log(n)$. But it is not bounded below, it can takes large negative values for the integers $n$ such that $\alpha^n$ is close to $1$.

I am surprised by the question. The sums are bounded if $\alpha \ne 1$, since for every $n \ge 1$, $$\Big|\sum_{k=1}^n \Re R_\alpha^k z \Big| = \Big|\Re \Big(\sum_{k=1}^n R_\alpha^k z \Big) \Big| = \Big| \Re \Big(\frac{\alpha-\alpha^{n+1}}{1-\alpha}z\Big) \Big| \le \Big|\frac{\alpha-\alpha^{n+1}}{1-\alpha}z\Big| \le \frac{2}{|1-\alpha|}.$$ The log of this quantity is bounded above, so it can not increase like $(1/2)\log(n)$. But it is not bounded below, it can takes large negative values for the integers $n$ such that $\alpha^n$ is close to $1$.

ADDENDUM : one can check that $$\limsup_{n \to +\infty} \log\Big|\sum_{k=1}^n \Re R_\alpha^k z \Big| \times \frac{1}{\log(n)} = 0,$$ whereas $$\liminf_{n \to +\infty} \log\Big|\sum_{k=1}^n \Re R_\alpha^k z \Big| \times \frac{1}{\log(n)} \le -1.$$ Indeed, setting $\alpha = e^{i\theta}$ with $\theta \in \mathbb{R}$, Dirichlet principle ensures that $|\alpha^n-1| = |e^{in\theta}-1| \le \mathrm{dist}(n\theta,\mathbb{Z}) < 1/n$ for infinitely many $n$. Actually, the liminf above is $-1$ for almost every $\alpha$.

I am surprised by the question. My remarks are too long for a comment. The sums are bounded if $\alpha \ne 1$, since for every $n \ge 1$, $$\Big|\sum_{k=1}^n \Re R_\alpha^k z \Big| = \Big|\Re \Big(\sum_{k=1}^n R_\alpha^k z \Big) \Big| = \Big| \Re \Big(\frac{\alpha-\alpha^{n+1}}{1-\alpha}z\Big) \Big| \le \Big|\frac{\alpha-\alpha^{n+1}}{1-\alpha}z\Big| \le \frac{2}{|1-\alpha|}.$$

Then, what is the definition of the logarithm of a negative number? Whatever the choice done for the imaginary part of the logarithm (i.e. the argument) the real part of the real part of the logarithm of the sums is bounded above by some constant, so you cannot find a limit $1/2$.

I am surprised by the question. The sums are bounded if $\alpha \ne 1$, since for every $n \ge 1$, $$\Big|\sum_{k=1}^n \Re R_\alpha^k z \Big| = \Big|\Re \Big(\sum_{k=1}^n R_\alpha^k z \Big) \Big| = \Big| \Re \Big(\frac{\alpha-\alpha^{n+1}}{1-\alpha}z\Big) \Big| \le \Big|\frac{\alpha-\alpha^{n+1}}{1-\alpha}z\Big| \le \frac{2}{|1-\alpha|}.$$ The log of this quantity is bounded above, so it can not increase like $(1/2)\log(n)$. But it is not bounded below, it can takes large negative values for the integers $n$ such that $\alpha^n$ is close to $1$.

I am surprised by the question. My remarks are too long for a comment. The sums are bounded if $\alpha \ne 1$, since for every $n \ge 1$, $$\Big|\sum_{k=1}^n \Re R_\alpha^k z \Big| = \Big|\Re \Big(\sum_{k=1}^n R_\alpha^k z \Big) \Big| = \Big| \Re \Big(\frac{\alpha-\alpha^{n+1}}{1-\alpha}z\Big) \Big| \le \Big|\frac{\alpha-\alpha^{n+1}}{1-\alpha}z\Big| \le \frac{2}{|1-\alpha|}.$$

Then, what is the definition of the logarithm of a complex number? Whatever the choice done for the imaginary part of the logarithm (i.e. the argument) the real part of the real part of the logarithm of the sums is bounded above by some constant, so you cannot find a limit $1/2$.

I am surprised by the question. My remarks are too long for a comment. The sums are bounded if $\alpha \ne 1$, since for every $n \ge 1$, $$\Big|\sum_{k=1}^n \Re R_\alpha^k z \Big| = \Big|\Re \Big(\sum_{k=1}^n R_\alpha^k z \Big) \Big| = \Big| \Re \Big(\frac{\alpha-\alpha^{n+1}}{1-\alpha}z\Big) \Big| \le \Big|\frac{\alpha-\alpha^{n+1}}{1-\alpha}z\Big| \le \frac{2}{|1-\alpha|}.$$

Then, what is the definition of the logarithm of a negative number? Whatever the choice done for the imaginary part of the logarithm (i.e. the argument) the real part of the real part of the logarithm of the sums is bounded above by some constant, so you cannot find a limit $1/2$.