determine whether sum converges or diverges: $\sum ^\infty_{n=2} \frac{\cos(\frac{\pi n^2}{n+1})}{\log^2n}$

Question

determine whether sum converges or diverges: $\sum ^\infty_{n=2} \frac{\cos(\frac{\pi n^2}{n+1})}{\log^2n}$

tried to use regular convergence tests, no success so far.

Answer 1

A rigorous proof could be the following:

1. $$ \begin{gathered} \cos \left( {\frac{{\pi n^2 }} {{n + 1}}} \right) = \cos \left[ {\pi \left( {n - 1} \right) + \frac{\pi } {{n + 1}}} \right] = \hfill \\ \hfill \\ = \cos \left[ {\pi \left( {n - 1} \right)} \right]\cos \left( {\frac{\pi } {{n + 1}}} \right) - \sin \left[ {\pi \left( {n - 1} \right)} \right]\sin \left( {\frac{\pi } {{n + 1}}} \right) = \hfill \\ \hfill \\ = \left( { - 1} \right)^{n - 1} \cos \left( {\frac{\pi } {{n + 1}}} \right) \hfill \\ \end{gathered} $$
2. $$ \sum\limits_{n = 2}^{ + \infty } {\frac{{\cos \left( {\frac{{\pi n^2 }} {{n + 1}}} \right)}} {{\log ^2 n}}} = \sum\limits_{n = 2}^{ + \infty } {\left( { - 1} \right)^{n - 1} \frac{{\cos \left( {\frac{\pi } {{n + 1}}} \right)}} {{\log ^2 n}}} $$
3. We prove now that $$ a_n = \frac{{\cos \left( {\frac{\pi } {{n + 1}}} \right)}} {{\log ^2 n}} \downarrow 0\,\,\,\,\,\,n \to + \infty $$
4. Let be $$ f(x) = \frac{{\cos \left( {\frac{\pi } {{x + 1}}} \right)}} {{\log ^2 x}} $$ with $x \geq 2$. Then $$ f'(x) = - \frac{{2\cos \left( {\frac{\pi } {{x + 1}}} \right)}} {{x\log ^3 (x)}} + \frac{{\pi \sin \left( {\frac{\pi } {{x + 1}}} \right)}} {{(x + 1)^2 \log ^2 (x)}} $$ thus $$ f'(x) = - \frac{{2\cos \left( {\frac{\pi } {{x + 1}}} \right)}} {{x\log ^3 (x)}} + o\left( {\frac{1} {{x\log ^3 (x)}}} \right) $$ as $x \to +\infty$. This means that eventually $f'(x)<0$ and therefore $f(x)$ is monotonically decreasing as well as $a_n$. Now the series is convergent by Leibniz Test.

Answer 2

well it's as simple as using a theorem. This is an alternating series which monotonically converges to 0. So according to Leibniz it converges. thanks to @lulu