Limit of a product as $N\rightarrow\infty$ with terms that depend on $N$ How do I find the find the limit of these functions as $N\rightarrow\infty$? I tried many things, trying to make them integrals, Taylor expansions, et cetera. Even a proof of existence of limit would be helpful, but an exact evaluation of the limit would be brilliant.
$p_N\left(z\right)=\prod\limits_{n=1}^{\left\lfloor\frac N2\right\rfloor}\left(1+e^{-2Nz\sin\left(\frac{\pi n}N\right)}\right)^2$
$q_N\left(z\right)=\prod\limits_{n=1}^{\left\lfloor\frac N2\right\rfloor}\left(1+e^{-2Nz\sin\left(\frac{\pi n}N-\frac\pi{2N}\right)}\right)^2$
$r_N\left(z\right)=\prod\limits_{n=1}^{\left\lfloor\frac N2\right\rfloor}\left(1-e^{-2Nz\sin\left(\frac{\pi n}N-\frac\pi{2N}\right)}\right)^2$
Here, $z\in\mathbb C$ and $\mathrm{Re}\left(z\right)>0$. I tried plotting these function for $N=10000$ and I found that they are periodic with periods $2$, $4$ and $4$ for $p$, $q$ and $r$ respectively. The first graph at $z=\frac{1+it}2$ and second is at $z=\frac{1}{20}+\frac{it}{2}$. Both are potted against $t$ for $0\leq t\leq16$.
Plot for $z=\frac{1+it}2$

Plot for $z=\frac{\frac{1}{10}+it}2$

 A: I am doing it for $p_N$, or more accurately, $a_N$ which is the same without the squaring.
Let $e_{n,N}=\exp{-2Nz\sin{\frac{n\pi}{N}}}$, and $r_{n,N}=\exp{-2Ns\sin{\frac{n\pi}{N}}}=|e_{n,N}|$ for $1 \leq n \leq N/2$, where $s$ is the real part of $z$. 
Note that $e_{n,N} \rightarrow e^{-2nz\pi}$ as $N \rightarrow \infty$. 
Let $n_0>0$, let $N > 2n_0$. 
Then $$\left|\frac{a_N(z)}{\prod_{n=1}^{n_0}{(1+e_{n,N})}}-1\right| \leq \left|\prod_{N/2 \geq n \geq n_0+1}{(1+e_{n,N})}-1 \right| \leq \left(\prod_{N/2 \geq n \geq n_0+1}{(1+r_{n,N})}\right)-1 \leq \exp\left(\sum\limits_{N/2 \geq n \geq n_0+1}{r_{n,N}}\right)-1.$$
As a consequence, since $r_{n,N} \leq e^{-4ns}$, hence
$$\left|\frac{a_N(z)}{\prod_{n=1}^{n_0}{(1+e_{n,N})}}-1\right| \leq \exp{\frac{e^{-4(n_0+1)s}}{1-e^{-4s}}}-1=f_{n_0}.$$
Thus $$\left|a_N(z)-\prod_{n=1}^{n_0}{(1+e_{n,N})}\right| \leq f_{n_0}\prod_{n=1}^{n_0}{(1+r_{n,N})} \leq f_{n_0}\exp{\sum_{n\geq 1}{e^{-4ns}}}=C(z)f_{n_0}.$$
Note that as a consequence, when $N$ goes to $\infty$, 
$$\lim \sup \left|a_N(z)-\prod_{n=1}^{n_0}{(1+e^{-2nz\pi})}\right| \leq C(z)f_{n_0}.$$
Since $f_{n_0}$ goes to $0$ and $\prod_{n=1}^{n_0}{(1+e^{-2nz\pi})}$ is absolutely convergent, 
$$p_N(z) \rightarrow \left(\prod_{n=1}^{\infty}{(1+e^{-2n\pi z})}\right)^2.$$
