Question

What is the radius of convergence?

What is the radius of convergence of `sum_(n=1)^in((3^(n+2))(x^n))/(n+1)`?

Answer 1

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

Explanation:

Apply the ratio test by evaluating:

`lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) abs ( ( (3^(n+3)x^(n+1))/(n+2)) / ( (3^(n+2)x^n)/(n+1)))`

`lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) (3^(n+3))/(3^(n+2)) (n+1)/(n+2) abs ( x^(n+1)/x^n)`

`lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) 3 absx (n+1)/(n+2)`

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

The series then is absolutely convergent for:

`3absx < 1`

`abs x < 1/3`

therefore the radius of convergence is `R=1/3`.