(A similar question I asked recently: Can anyone detemine the convergence of somewhat like p-series?) (Link for reference: Convergence or divergence of $\sum_{n=1}^{\infty}\frac{1}{n^{2-\cos n}}$)
Reference)We know that $\sum_{n=1}^{\infty}\frac{1}{n^{1}} = {\infty}$ & $\sum_{n=2}^{\infty}\frac{1}{nlogn} = {\infty}$ &
$\sum_{n=2}^{\infty}\frac{1}{prime(n)} = {\infty}$ ($prime(n)$ ~ ${nlogn}$ for large n) &
$\sum_{n=1}^{\infty}\frac{1}{n^{2}} = \frac{\pi^2}{{6}}$
$\sum_{n=2}^{\infty}\frac{1}{nlogn}$ and prime(n) is less denser than $\frac{1}{n}$ but dense enough to make the sum diverge. (prime(n) is the nth prime) And, thanks to reference link, we know that $\sum_{n=1}^{\infty}\frac{1}{n^{2-sin(n)}}$ also diverges, which implies 2-sin(n) gets close to 1 frequently enough to make the sum diverges.
Question. $\sum_{n=2}^{\infty}\frac{1}{{nlogn}^{2-sin(n)}}$ or, $\sum_{n=2}^{\infty}\frac{1}{{nlogn}^{2-sin(nlogn)}}$ or , $\sum_{n=2}^{\infty}\frac{1}{{prime(n)}^{2-sin(n)}}$ diverges? or converges?
