The first one you would argue that (log(n))^3 < (n^(1/9))^3 = n^(1/3) for large n, now since n^(1/3) as the numerator converges so does the series.
The second one you would do a comparison test with 1/n, eg [1/n^(1+1/n)]/[1/n] = 1 as n approaches infinity, and we know that if one series diverges both series diverges, so the original series diverges since 1/n diverges.
The third one is just like the first one I think, log(n!) < n^(3/2) for large n.