What is the radius of convergence?

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

1 Answer
Apr 4, 2018

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.