How do you find the radius of convergence #Sigma n^nx^n# from #n=[1,oo)#?

Question

14150 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-18T09:04:44

The radius of convergence is #R=0#, as the series

#sum_(n=1)^oo n^nx^n#

is divergent for any #x !=0#

Given the series:

#sum_(n=1)^oo n^nx^n#

we can apply the ratio test:

#abs (a_(n+1)/a_n) = (n+1)^(n+1)/n^n abs(x)^(n+1)/abs(x)^n = (n+1)((n+1)/n)^nabs(x)#

so that:

#lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) (n+1)(1+1/n)^nabs(x)#

as:

#lim_(n->oo) (1+1/n)^n = e#

for any #x>0# we have:

#lim_(n->oo) abs (a_(n+1)/a_n) = oo #

and the series is divergent, so the radius of convergence is #R=0#