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?


Let $$f(x)=\sum_{n=1}^{\infty}\dfrac{\sinh{(nx)}}{\sinh{(n\pi)}}$$ so $$f'(x)=\sum_{n=1}^{+\infty}\dfrac{n\cosh{(nx)}}{\sinh{(n\pi)}}$$ and other hand $$f(x)=\dfrac{1}{e^{\pi-x}-1}+\sum_{n=1}^{\infty}\dfrac{e^{-(\pi-x)n}(e^{-2n\pi}-e^{-2n\pi})}{1-e^{-2n\pi}}$$ so $$f'(x)=\dfrac{e^{\pi-x}}{(e^{\pi-x}-1)^2}+\sum_{n=1}^{\infty}\dfrac{ne^{-(\pi-x)n}(e^{-2n\pi}-e^{2nx})+2ne^{-(\pi-x)n}e^{-2nx}}{1-e^{-2n\pi}}$$ and it not easy to prove $$\lim_{x\to \pi^{-}}\sum_{n=1}^{\infty}\dfrac{ne^{-(\pi-x)n}(e^{-2n\pi}-e^{2nx})+2ne^{-(\pi-x)n}e^{-2nx}}{1-e^{-2n\pi}}=2\sum_{n=1}^{\infty}\dfrac{n}{e^{2n\pi}-1}=\dfrac{1}{12}-\dfrac{1}{4\pi}$$ so you can prove by done


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