Prolem: ${\mathop {\lim }\limits_{x \to {\pi ^ - }} \mathop {\lim }\limits_{n \to \infty } \left( {{n^2}\left( {1 - \sin \frac{\pi }{{2n}} \cdot \sum\limits_{k = 1}^{n - 1} {\sin \left( {\frac{{\left( {2\left\lfloor {\sqrt {kn} } \right\rfloor + 1} \right)\pi }}{{2n}}} \right)} } \right)\left( {\frac{1}{{4{{\cos }^2}\frac{x}{2}}} - \sum\limits_{n = 1}^\infty {\frac{{n\cosh \left( {nx} \right)}}{{\sinh \left( {n\pi } \right)}}} - \frac{1}{{12}}} \right)} \right)}$,
where $\lfloor\cdot\rfloor$ is greatest integer function.
Add: In my opinion, we must solve $$\mathop {\lim }\limits_{n \to \infty } {n^2}\left( {1 - \sin \frac{\pi }{{2n}} \cdot \sum\limits_{k = 1}^{n - 1} {\sin \left( {\frac{{\left( {2\left\lfloor {\sqrt {kn} } \right\rfloor + 1} \right)\pi }}{{2n}}} \right)} } \right)$$ and $$\mathop {\lim }\limits_{x \to {\pi ^ - }} \left( {\frac{1}{{4{{\cos }^2}\frac{x}{2}}} - \sum\limits_{n = 1}^\infty {\frac{{n\cosh \left( {nx} \right)}}{{\sinh \left( {n\pi } \right)}}} - \frac{1}{{12}}} \right).$$ But how can we obtain them?
And in this, we have $$\sum\limits_{k = 1}^{n - 1} {\sin \left( {\frac{{\left( {2\left\lfloor {\sqrt {kn} } \right\rfloor + 1} \right)\pi }}{{2n}}} \right)} = \cot \left( {\frac{\pi }{{2n}}} \right)\cos \left( {\frac{\pi }{{2n}}} \right).$$ So we have \begin{align*} &\mathop {\lim }\limits_{n \to \infty } {n^2}\left( {1 - \sin \frac{\pi }{{2n}} \cdot \sum\limits_{k = 1}^{n - 1} {\sin \left( {\frac{{\left( {2\left\lfloor {\sqrt {kn} } \right\rfloor + 1} \right)\pi }}{{2n}}} \right)} } \right) \\ =& \mathop {\lim }\limits_{n \to \infty } {n^2}\left( {1 - \sin \frac{\pi }{{2n}} \cdot \cot \left( {\frac{\pi }{{2n}}} \right)\cos \left( {\frac{\pi }{{2n}}} \right)} \right)\\ = &\mathop {\lim }\limits_{n \to \infty } {n^2}\left( {1 - {{\cos }^2}\left( {\frac{\pi }{{2n}}} \right)} \right) = \mathop {\lim }\limits_{n \to \infty } {n^2}{\sin ^2}\left( {\frac{\pi }{{2n}}} \right) = \frac{{{\pi ^2}}}{4} \end{align*}
What about the next one? How to solve it?