How do you find the radius of convergence #Sigma (n^nx^n)/(lnn)^n# from #n=[2,oo)#?

1 Answer
mason m
Mar 1, 2017

The root test states that the series #suma_n# is convergent if #L=lim_(nrarroo)root(n)abs(a_n)<1#.

Here we see that #a_n=((nx)/lnn)^n#, so:

#L=lim_(nrarroo)root(n)abs(((nx)/lnn)^n)=lim_(nrarroo)abs((xn)/lnn)#

Since the limit is dependent only on the change in #n#, the #x# can be removed from the limit like a constant:

#L=absxlim_(nrarroo)abs(n/lnn)#

We see that the limit approaches #oo#, since #n# grows faster than the logarithmic function #lnn#. The only time when #L<1#, when the series converges, will be when #L=0#, which occurs only when #x=0#.

Since there is only one value of #x# for which the series converges, we see that #R=0#.

Related questions
See all questions in Determining the Radius and Interval of Convergence for a Power Series
Impact of this question
1478 views around the world
You can reuse this answer
Creative Commons License