Sequences undergrad

Let $u=\log x$, then $du=\frac{dx}{x}$. So 
\[\int \frac{1}{x \log ^{8/5}x}=\int \frac{1}{u^{8/5}}du=\int u^{-8/5}du=-\frac{3}{5}u^{-\frac{3}{5}}+c\]
\[=-\frac{5}{3}(\log x)^{-\frac{3}{5}}+c\]
Note that $-8/5+1=-3/5$.

Note also that 
\[\int_2^{\infty} \frac{1}{x \log ^{8/5}x}=(-\frac{5}{3}(\log x)^{-\frac{3}{5}})|_2^{\infty}=0+\frac{5}{3}(\log 2)^{-\frac{3}{5}}<\infty,\]
since the above integral is finite, by the integral test, the infinite series
\[\sum_{n=2}^{\infty}\frac{1}{n \log ^{8/5}n}\]
is also convergenet.