Question

How do you test the series `Sigma rootn(n)/n` from n is `[1,oo)` for convergence?

Answer 1

`sum_(n=1)^oo root(n)(n)/n` diverges.

Explanation:

`sum_(n=1)^oo root(n)(n)/n > sum_(n=1)^oo1/n` and `sum_(n=1)^oo1/n` diverges so `sum_(n=1)^oo root(n)(n)/n` diverges also.