Does there exist $f(n)>n\ln n$ such that $$\sum_{n=k}^\infty\frac{1}{f(n)}$$ diverges?
If so, is there a maximum such $f(n)$?
Of course the answer to the last question I suppose would eliminate the need for all convergence tests (given that the series is not alternating)!
By Cauchy's condensation test, all the following series are divergent: $$ \sum_{n\geq n_0}\frac{1}{n\log(n)},\qquad \sum_{n\geq n_0}\frac{1}{n\log(n)\log\log(n)},\qquad \sum_{n\geq n_0}\frac{1}{n\log(n)\log\log(n)\log\log\log(n)} $$ and all the following series are convergent: $$ \sum_{n\geq n_0}\frac{1}{n\log(n)^2},\qquad\! \sum_{n\geq n_0}\frac{1}{n\log(n)\log\log(n)^2},\qquad \!\sum_{n\geq n_0}\frac{1}{n\log(n)\log\log(n)\log\log\log(n)^2} $$ so there is clearly no maximum.
In each case, $n_0$ is chosen in such a way that every term of the corresponding sum makes sense.