Is square of Poisson kernel conditionally convergent?

Question

Poisson kernel is defined for $\theta\in [0,2\pi)$ as $$P_r(\theta) = \sum\limits_{n\in\mathbb{Z}} r^{|n|}e^{in\theta}$$ which equals $$P_r(\theta) = \frac{1-r^2}{r^2-2r\cos (\theta) + 1 }.$$ For this second expression, clearly for all $\theta\neq 0, \ P_r^2(\theta)$ tends to $0$ as $r$ tends to $1$. But calculating the same for the first expression gives \begin{eqnarray} &P_r(\theta)&=\sum\limits_{n_1\in\mathbb{Z}}r_1^{|n_1|}e^{in_1\theta}\sum\limits_{n_2\in\mathbb{Z}}r_2^{|n_2|}e^{in_1\theta} \\&&=\sum\limits_{n_1\in\mathbb{Z}}\sum\limits_{n_2\in\mathbb{Z}}r^{|n_1|+|n_2|}e^{i(n_1+n_2)\theta} \end{eqnarray} Let $\theta\neq 0$ and observe $$\forall n\in\mathbb{N}, \ r^{|n|+|-n|}e^{i(n-n)\theta}=r^{|2n|}.$$ Therefore \begin{eqnarray} &P_r^2(\theta )&=\sum\limits_{(n_1,n_2)\in\mathbb{Z}\times\mathbb{Z}}r^{|n_1|+|n_2|}e^{i(n_1+n_2)\theta} \\&&=\sum\limits_{ (n_1,n_2)\in\mathbb{Z}\times\mathbb{Z} \\ n_1\neq - n_2 }r^{|n_1|+|n_2|}e^{i(n_1+n2)\theta} + \sum\limits_{n\in\mathbb{Z}}r^{|2n|}e^{i(n-n)\theta} \\&&=\sum\limits_{ (n_1,n_2)\in\mathbb{Z}\times\mathbb{Z} \\ n_1\neq - n_2 }r^{|n_1|+|n_2|}e^{i(n_1+n2)\theta} + \sum\limits_{n\in\mathbb{Z}}r^{|2n|} \end{eqnarray} In the last line the second term is a geometric sequence and converges to $2/(1-r^2)-1$, which diverges (for all $\theta$) as $r$ tends to $1$. This makes me think Is the square of Poisson kernel conditionally convergent?