I have already proven that $$\sum_{n=2}^{\infty}\frac{\cos(\pi n)}{\ln(n)\cdot \ln(n^n + n)}$$ conditionally converge and $$\sum_{n=2}^{\infty} \frac{n\cdot \cos(2n)}{n^2 + 11}$$ converge, hence $$\sum_{n=2}^{\infty} \frac{n\cdot \cos(2n)}{n^2 + 11} + \frac{\cos(\pi n)}{\ln(n)\cdot \ln(n^n + n)}$$ converge, but I wasn't able to determine whether it is conditionally or absolutely.
If I can prove that $$\sum_{n=2}^{\infty} \frac{n\cdot \cos(2n)}{n^2 + 11}$$ converge absolutely, then using the triangle inequality I get $$-\sum_{n=2}^{\infty}\left|\frac{\cos(\pi n)}{\ln(n)\cdot \ln(n^n + n)}\right| \le \sum_{n=2}^{\infty}\left|\frac{n\cdot \cos(2n)}{n^2 + 11}\right| - \sum_{n=2}^{\infty} \left|\frac{n\cdot \cos(2n)}{n^2 + 11} + \frac{\cos(\pi n)}{\ln(n)\cdot \ln(n^n + n)}\right|$$ which means if $\sum_{n=2}^{\infty}\left|\frac{n\cdot \cos(2n)}{n^2 + 11} + \frac{\cos(\pi n)}{\ln(n)\cdot \ln(n^n + n)}\right|$ converged I would have reached a contradiction. But I wasn't able to find a way to prove that $\sum_{n=2}^{\infty} \frac{n\cdot \cos(2n)}{n^2 + 11}$ converge absolutely nor that it conditionally converge, in which case my attempt to prove this won't work
$\sum_{n = 2}^{\infty} \left \lvert \frac{n \cdot cos(2 \cdot n)}{n^{2} + 11} \right \rvert \sim $(in terms of convergence) $\sum_{n = 2}^{\infty} \left \lvert \frac{n \cdot cos(2 \cdot n)}{n^{2}} \right \rvert = \sum_{n = 2}^{\infty} \left \lvert \frac{cos(2 \cdot n)}{n} \right \rvert$. And $\sum_{n = 2}^{\infty} \left \lvert \frac{cos(2 \cdot n)}{n} \right \rvert > \sum_{n = 2}^{\infty} \frac {cos^2(2 \cdot n)}{n} = \sum_{n = 2}^{\infty} \frac {1 - cos(4 \cdot n)}{2 \cdot n}$, which, obviously diverges, because $\sum_{n = 2}^{\infty} \frac{1}{n}$ diverges. So $\sum_{n = 2}^{\infty} \frac{n \cdot cos(2 \cdot n)}{n^{2} + 11}$ converges conditionally. Howewer, $\lvert cos(\pi \cdot n) \rvert = 1$ for natural n, and $\sum_{n = 4}^{\infty} \frac{1}{ln(n) * ln(n^{n} + n)}$ converges, so $\sum_{n = 4}^{\infty} \frac{cos(\pi \cdot n)}{ln(n) * ln(n^{n} + n)}$ converges absoulutely. So the series in your question converge conditionally, because $\sum_{n = 2}^{\infty} \frac{n \cdot cos(2 \cdot n)}{n^{2} + 11}$ converges conditionally, $\sum_{n = 4}^{\infty} \frac{cos(\pi \cdot n)}{ln(n) * ln(n^{n} + n)}$ converges absoulutely.