Circular sequences continuous? I noticed something interesting when playing around with Mathematica.
Consider the sum
$$x(N)= \frac{1}{N^2} \sum_{i=1}^{N-1} \frac{1}{1-\cos(2\pi i/N)}$$
this sequence will converge to $1/6$ as $N$ tends to infinity.
Now, if I look at 
$$y(N)= \frac{1}{N^2} \sum_{i=1}^{N-1} \frac{1}{1-\cos((2\pi i+\pi)/N)}$$
this sequence will no longer converge to $1/6$, yet we get, according to mathematica, something close to like $0.297358.$
Now, I got curious and tried to see whether the limiting expression is continuous in its argument, i.e. I wanted to start with something that for $N$ small is more like $y(N)$ and whose summands tend pointwise to the summands of $x(N)$ as $N$ tends to infinity
$$z_{\alpha}(N)= \frac{1}{N^2} \sum_{i=1}^{N-1} \frac{1}{1-\cos((2\pi i+\pi/N^{\alpha})/N)}.$$
From Mathematica I seem to get that if $\alpha=1$ then $z_{1}(N)$ tends to $1/6$. The same is true if $\alpha=1/4$, for example. Then, I tried something like $\alpha=1/100$ and it was no longer clear whether it would converge. 
Therefore, my question is: Does $z_{\alpha}(N)$ tends to $1/6$ for all $\alpha>0$ or is there some threshold at which it stops converging? Is the convergence with some explicit rate?
Please let me know if you have any questions.
 A: Claim: The sequence $z_{\alpha}(N)$ tends to $1/6$ for all $\alpha>0$, and the error is
 $O(N^{-\min\{1,\alpha\}})$. Let's assume $\alpha \le 1$. For fixed $\alpha$ and $1 \le k<N$, write
$$u_N(k):=\frac{1}{N^2}  \cdot  \frac{1}{1-\cos((2\pi k+\pi/N^{\alpha})/N)} \,  .$$
 It suffices to prove that (ignoring unimportant floor and ceiling symbols below)
$$w_{\alpha}(N):=   \sum_{k=1}^{N/7} u_N(k)$$
satisfies 
$$|w_{\alpha}(N)-1/12| =O(1/N) \,. \quad (*)$$
 Indeed, since $1-\cos(2\pi x)$ has a positive lower bound on $[1/7,6/7]$, the sum $\sum_{k= N/7+1}^{6N/7-1} u_N(k)$ is $O(1/N)$ and the sum $\sum_{k= 6N/7}^{N-1} u_N(k)$
can be handled in the same way as $w_{\alpha}(N)$ using $\cos(x)=\cos(2\pi-x)$.
Proof of (*):
Taylor expansion of cosine gives $|1-\cos(x)-x^2/2| \le x^4/12$ for $x \in [0,1]$,
so for $1 \le k<N/7$ $$|1-\cos((2\pi k+\pi/N^{\alpha})/N)-(2\pi k+\pi/N^{\alpha})^2/(2N^2)| \le 10 \cdot 2\pi^2 \cdot k^4/N^4 \,.$$
Thus 
$$w_{\alpha}(N)=   \frac{1}{2\pi^2} \sum_{k=1}^{N/7} \frac{1}{  (k+1/(2N^{\alpha}))^2+a_k k^4/N^2} $$
where $|a_k| \le 10$. Since $|1/x-1/k^2| \le 2|x-k^2|/k^4$ for $x>k^2/2$, we deduce that 
$$w_{\alpha}(N)=   \frac{1}{2\pi^2} \sum_{k=1}^{N/7} \Bigl(\frac{1}{k^2}+O(N^{-\alpha}/k^3)+O(N^{-2}) \Bigr) = \frac{1}{2\pi^2}\Bigl(\pi^2/6+O(N^{-\alpha})\Bigr)\,.   $$
This proves (*). 
The same method shows that 
$$y(N):= \frac{1}{N^2} \sum_{k=1}^{N-1} \frac{1}{1-\cos((2\pi k+\pi)/N)}$$
converges to 
$$ \frac{2}{\pi^2} \Bigl(\sum_{k \ge 1} \frac{1}{(2k+1)^2}+ \sum_{k \ge 1}\frac{1}{(2k-1)^2} \Bigr)=\frac{2}{\pi^2}\Bigl(2\frac{\pi^2}{8}-1\Bigr)=1/2-2/\pi^2=0.2973576327153...$$
