Question

What are the values of r (with r>0) for which the series converges?

Show that `r^ln(n)=n^ln(r)`. Then determine the values of r (with `r>0`) for which the series `sum_(n=1)^oor^ln(n)` converges.

Answer 1

`r<1/e` is the condition for the convergence of `sum_(n=1)^oor^ln(n)`

Explanation:

I will just answer the part about the convergence, the first part having been answered in the comments. We can use `r^ln(n)=n^ln(r)` to rewrite the sum `sum_(n=1)^oor^ln(n)` in the form

`sum_(n=1)^oon^ln(r) = sum_(n=1)^oo 1/n^p,qquad mbox{ for }p=-ln(r)`

The series on the right is the series form for the famous Riemann Zeta function. It is well known that this series converges when `p>1`. Using this result directly gives

`-ln(r)>1 implies ln(r)<-1 implies r < e^-1 = 1/e`

The result about the Riemann Zeta functions is very well known, If you want an ab initio answer, you can try the integral test for convergence .