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`.