Question

How do you find the radius of convergence `Sigma 3^nx^(2n)` from `n=[1,oo)`?

Answer 1

`1/sqrt3`

Explanation:

`sum_(n=1)^oo3^nx^(2n)`

Recall that the series `sum_(n=1)^ooa_n` converges if `lim_(nrarroo)abs(a_(n+1)/a_n)<1` through the ratio test.

We can take `a_n=3^nx^(2n)` and apply the ratio test and find the values of `x` for which the properties of convergence are satisfied (when the absolute value of the ratio is less than `1`).

`lim_(nrarroo)abs(a_(n+1)/a_n)=lim_(nrarroo)abs((3^(n+1)x^(2(n+1)))/(3^nx^(2n)))<1`

`color(white)(lim_(nrarroo)abs(a_(n+1)/a_n)=)lim_(nrarroo)abs((3^(n+1)x^(2n+2))/(3^nx^(2n)))<1`

`color(white)(lim_(nrarroo)abs(a_(n+1)/a_n)=)lim_(nrarroo)abs(3x^2)<1`

`n` is no longer a part of the limit, so we can drop that part. Furthermore, `3x^2>0` for all Real values of `x`, so the absolute value bars are unnecessary.

`color(white)(lim_(nrarroo)abs(a_(n+1)/a_n)=)3x^2-1<0`

`color(white)(lim_(nrarroo)abs(a_(n+1)/a_n)=)(sqrt3x+1)(sqrt3x-1)<0`

This is true on `-1/sqrt3 < x < 1/sqrt3`.

Since this is the interval of convergence (for this problem we don't need to check the bounds), we see that the radius of convergence `R` is `1/sqrt3`.