testing for the convergence of a series I have to check if the following series converges. I thought that I have to use comparison test, but it did not lead me anywhere.
$$\sum_{n=2}^{\infty}\frac{ln^5(2n^7+13)+10sin(n)}{nln^6(n^{7/8}+2n^{1/2}-1)ln(ln(n+(-1)^n))}$$

Note that for sufficiently large $n$
$$\frac{\ln^5{(2n^7+13)+10\sin(n)}}{\ln^5{(n^\frac{7}{8}}+2\sqrt{n}-1)} > \frac{\ln^5{(2n^7+13)-10}}{\ln^5{(n^\frac{7}{8}}+2\sqrt{n}-1)} > 1.$$
Also 
$$ \lim_{n \to \infty}\frac{n^{\frac{7}{8}}+2\sqrt{n}-1}{n} = 0,$$
and for sufficiently large $n$ we have
$$ \ln(n^{\frac{7}{8}}+2\sqrt{n}-1) < \ln n.$$
Since, $ n+ (-1)^n \leqslant n+1$, we have $\ln \ln (n+ (-1)^n) \leqslant \ln \ln (n+1)$.
Thus, for sufficiently large $n$,
$$\frac{\ln^5{(2n^7+13)+10\sin(n)}}{n\cdot \ln^6{(n^\frac{7}{8}}+2\sqrt{n}-1)\cdot\ln{\ln{(n+(-1)^n}})} > \frac{1}{n \cdot \ln n \cdot \ln \ln (n+1)} \\ > \frac{1}{(n+1) \cdot \ln (n+1) \cdot \ln \ln (n+1)} .$$
The series with terms given by the RHS of the above inequality diverges by the integral test, since
$$\int_2^{\infty}\frac{dx}{(x+1) \cdot \ln (x+1) \cdot \ln \ln (x+1) }= \lim_{x \to \infty}[\ln \ln \ln(x+1)- \ln \ln \ln (3)] = \infty.$$
