use the ratio test:
[note: for the series sum_(n=1)^ooa_n, if lim_(nrarroo)abs(a_(n+1)/a_n)
-is <1, series converges
-is =1, inconclusive
-is >1, series diverges]
if you apply the ratio test to this: sum_(n=1)^oo(x-5)^n/(n4^n), evaluating lim_(nrarroo)abs(((x-5)^(n+1)/((n+1)(4^(n+1))))/((x-5)^(n)/((n)(4^(n)))) will show what x-values make the series converge or diverge.
simplifying the limit: =lim_(nrarroo)abs(((x-5)/((n+1)(4)))/(1/((n)))
=lim_(nrarroo)abs((n(x-5))/(4n+4))
[note: here you can divide both the numerator and denominator by n, because for whatever n value is used, the value inside the absolute value signs will stay the same (dividing by n/n or 1)]
=lim_(nrarroo)abs(((n(x-5))/n)/((4n)/n+4/n))
=abs((x-5)/(4+0))
=abs((x-5)/4)
back to the ratio test, the series can only converge if abs((x-5)/4)<1 or abs((x-5)/4)=1
Case 1: abs(x-5)<4
-4<x-5<4
1<x<9 (the solution interval must include these values)
Case 2: abs((x-5)/4)=1
x-5=-4, x-5=4
x=1,9
if x=1, series becomes: sum_(n=1)^oo(1-5)^n/(n4^n)
=sum_(n=1)^oo(-4)^n/(n4^n)
=sum_(n=1)^oo(-1)^n/(n)
this is the alternating harmonic series, which converges by the alternating series test
if x=9, series becomes: sum_(n=1)^oo(9-5)^n/(n4^n)
=sum_(n=1)^oo(4)^n/(n4^n)
=sum_(n=1)^oo1/n
this is the harmonic series, which diverges. here is a proof
so include x=1 in the interval, too: 1<=x<9
radius of convergence is half the difference between the upper and lower values for the interval =(9-1)/2=4
and here is a video with a similar problem
