Question

How do you find the radius of convergence `Sigma (1*4*7* * * (3n+1))/(n!)x^n` from `n=[0,oo)`?

Answer 1

The radius of convergence is `R= 1/3`

Explanation:

The ratio test states that a necessary condition for `sum_(n=0)^oo a_n` to converge is that:

`L = lim_(n->oo) abs(a_(n+1)/a_n) <= 1`

if `L<1` the condition is also sufficient and the series converges absolutely, while for `L=1` the test is inconclusive.

We then evaluate the ratio for the series at hand:

`abs(a_(n+1)/a_n) = (abs(x)^(n+1)(prod_1^(n+1) (3k+1))/((n+1)!))/(abs(x)^n(prod_1^n (3k+1))/(n!))= abs(x)(3(n+1) +1)/(n+1) = abs(x) (3n+4)/(n+1)`

so that:

`lim_(n->oo) abs(a_(n+1)/a_n) = 3abs(x)`

The series is then absolutely convergent for `absx <1/3` and divergent for `absx >1/3`with means that the radius of convergence is `R= 1/3`