Question

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

Answer 1

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

Explanation:

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`