When is a particular sum $\Theta(n)$? Define
$$S_n = \prod_{x=1}^{\lceil\frac{n}{\ln{n} }\rceil} \left(\frac{1}{\sqrt{n}} + \frac{2x}{n}\left(z_n-\frac{1}{\sqrt{n}} \right)\right) .$$

I am trying to work out necessary and sufficient conditions for a simply stated, positive,
  non-increasing function $z_n$  with $1/\sqrt{n} \leq z_n \leq 1/2$ so that
$$- \ln{S_n} = \Theta(n).$$

Clearly $z_n = 1/\sqrt{n}$ works and I know that $z_n = 1/2$ does not work for example.
Here is my attempt so far $$\prod_{x=1}^{k} (A + Bx) = \frac{B^k \Gamma(k+1+A/B)}{\Gamma(1+A/B)}.$$
In our case $A = 1/\sqrt{n}$ and $B= (2\sqrt{n} z_n - n)/n^{3/2}$ and 
$$A/B = \frac{n}{2\sqrt{n} z_n - n}$$  
We know from Asymptotics of $\prod_{x=1}^{\lceil\frac{n}{\log_2{n} }\rceil} \left(\frac{1}{\sqrt{n}} + x\left(\frac{1}{n}-\frac{2}{n^\frac{3}{2}} \right)\right) $ that
\begin{align}
\ln S_n&=k\ln B+\ln \Gamma(1+k+A/B)-\ln \Gamma(1+A/B)\approx\\
&\approx k\ln B+ \left(k+\frac AB\right)\left[\ln\left(k+\frac AB\right)-1\right]+\frac12\ln\left(k+\frac AB\right)\\&\qquad -\frac AB\left[\ln\frac AB -1\right]-\frac12\ln\frac AB\\
&\approx \color{red}{k\ln B}+\left(\color{red}{k}+\frac AB\right)\left[\color{red}{\ln k} +\frac{A}{Bk}-\frac12\left(\frac{A}{Bk}\right)^2-\color{red}{1}\right]+\frac12\ln k\\
&\qquad -\frac AB\left[\ln\frac AB -1\right]-\frac12\ln\frac AB.
\end{align}
It feels like with the right log approximation this might be close but I am not sure where to take it from here.
 A: Note that
$$
\prod_{x=1}^{\lceil n/\ln n \rceil} \max\bigg\{ \frac{1}{\sqrt{n}},  \frac{2x}{n}\bigg(z_n-\frac{1}{\sqrt{n}} \bigg)\bigg\} \le S_n \le \prod_{x=1}^{\lceil n/\ln n \rceil} 2\max\bigg\{ \frac{1}{\sqrt{n}},  \frac{2x}{n}\bigg(z_n-\frac{1}{\sqrt{n}} \bigg)\bigg\},
$$
and so
\begin{align*}
-\ln S_n &= \sum_{x=1}^{\lceil n/\ln n \rceil} \min\bigg\{ \tfrac12\ln n,  \ln \frac{n^{3/2}}{2x} - \ln (z_n\sqrt n-1)\bigg\} + O\bigg( \frac{n\ln 2}{\ln n} \bigg) \\
&= \sum_{1\le x\le \min\{n/2(z_n\sqrt n-1),\lceil n/\ln n \rceil\}} \tfrac12\ln n \\
&\qquad{}+ \sum_{n/2(z_n\sqrt n-1)<x\le\lceil n/\ln n \rceil} \bigg( \ln \frac{n^{3/2}}{2x} - \ln (z_n\sqrt n-1) \bigg) + O\bigg( \frac n{\ln n} \bigg).
\end{align*}
If $(z_n\sqrt n-1)/\ln n$ is bounded away from $0$, then the first term is already $\Omega(n)$. If $(z_n\sqrt n-1)/\ln n \to 0$, then the first term is $o(n)$ and so
$$
-\ln S_n = \sum_{n/2(z_n\sqrt n-1)<x\le\lceil n/\ln n \rceil} \bigg( \ln \frac{n^{3/2}}{2x} - \ln (z_n\sqrt n-1) \bigg) + o(n).
$$
Ran out of time - see if someone can take it from here....
