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.

1 Answer
Mar 10, 2018

Answer:

#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 .