How do you find the radius of convergence #Sigma (n^nx^n)/(ln(lnn))^n# from #n=[3,oo)#?

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Answer 1 · 2017-03-02T21:58:36

The interval of convergence is the single value #x=0# and the radius of convergence is #R=0#.

#sum_(n=3)^oo(n^nx^n)/(ln(lnn))^n=sum_(n=3)^oo((nx)/ln(lnn))^n#

The root test says that the series #suma_n# converges if #lim_(nrarroo)rootnabs(a_n)<1#.

So, we will take #L=lim_(nrarroo)rootnabs(a_n)# and find the values of #x# that make #L<1#, as this is when the series will converge.

#L=lim_(nrarroo)rootnabs(a_n)=lim_(nrarroo)rootnabs(((nx)/ln(lnn))^n)#

#color(white)L=lim_(nrarroo)abs((nx)/ln(lnn))#

Moving the #x# from the limit, since the limit only depends on the changing of #n#.

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

Since #n# grows faster than any logarithmic function, we see that #lim_(nrarroo)abs(n/ln(lnn))=oo#.

The only time, then, when #L<1# will be when the limit is multiplied by #0#, that is, when #absx=0#, which occurs only at #x=0#.

Thus, the interval of convergence is the single value #x=0# and the radius of convergence is #R=0#.