How do you find the interval of convergence #Sigma (3^n(x-4)^(2n))/n^2# from #n=[1,oo)#?

Where #a_n=(3^n(x-4)^(2n))/n^2#, find the ratio #a_(n+1)/a_n#:

#a_(n+1)/a_n=(3^(n+1)(x-4)^(2n+2))/(n+1)^2*n^2/(3^n(x-4)^(2n))#

#=3^(n+1)/3^n((x-4)^(2n+2)/(x-4)^(2n))n^2/(n+1)^2#

#=3(x-4)^2(n/(n+1))^2#

And:

#lim_(nrarroo)abs(a_(n+1)/a_n)=lim_(nrarroo)abs(3(x-4)^2(n/(n+1))^2)#

#=3(x-4)^2lim_(nrarroo)abs((n/(n+1))^2)#

The limit approaches #1#:

#=3(x-4)^2#

Through the ratio test, the series converges when #lim_(nrarroo)abs(a_(n+1)/a_n)<1#:

#3(x-4)^2<1#

#-1/sqrt3 < x-4 < 1/sqrt3#

#4-1/sqrt3 < x < 4+1/sqrt3#

Test the intervals to see if the integral converges at the extremes:

At #x=4-1/sqrt3#, the series is:

#sum_(n=1)^oo(3^n((4-1/sqrt3)-4)^(2n))/n^2=sum_(n=1)^oo(3^n(-1/sqrt3)^(2n))/n^2#

#=sum_(n=1)^oo(3^n(-1)^(2n)(1/sqrt3)^(2n))/n^2=sum_(n=1)^oo(3^n(1/3^n))/n^2=sum_(n=1)^oo1/n^2#

Which converges through the p-series test, so #x=4-1/sqrt3# is included in the interval of convergence.

At #x=4+1/sqrt3#:

#sum_(n=1)^oo(3^n((4+1/sqrt3)-4)^(2n))/n^2=sum_(n=1)^oo(3^n(1/sqrt3)^(2n))/n^2=sum_(n=1)^oo1/n^2#

Which converges as well. So the interval of convergence is:

#4-1/sqrt3 <= x <= 4+1/sqrt3#