How do you find the radius of convergence #Sigma (x^n)/(3^(n^2))# from #n=[0,oo)#?

Use the ratio test to evaluate the values of #x# for which the series is convergent:

#lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) abs ( (x^(n+1)/3^((n+1)^2))/(x^n/3^(n^2))#

#lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) abs ( x^(n+1)/x^n) 3^(n^2)/3^((n+1)^2) #

#lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) abs (x) 3^(n^2)/3^(n^2+2n+1) = abs x lim_(n->oo) 1/3^(2n+1) = 0#

So the series is absolutely convergent for every #x in RR#

We can also use the root test, which states that if #L=lim_(nrarroo)root(n)abs(a_n)<1#, then #sum_(n=0)^ooa_n# converges.

#L=lim_(nrarroo)root(n)abs(a_n)=lim_(nrarroo)(abs(x^n/3^(n^2)))^(1/n)#

Bringing the exponent in:

#L=lim_(nrarroo)abs(x^(n(1/n))/3^(n^2(1/n)))=lim_(nrarroo)abs(x/3^n)#

The limit is only dependent on #n#, so the #x# term can be moved from the limit.

#L=absxlim_(nrarroo)abs(1/3^n)#

As the denominator of the limit approaches #oo#, we see that the limit equals #0#. Thus,

#L=0#

We know that the series will diverge when #L<1#. Here, since #L=0#, this is always true, irrespective of the value of #x#.

Thus, the series converges on #-oo < x < oo# and #R=oo#.