Question

How do you find the interval of convergence `Sigma n3^nx^n` from `n=[0,oo)`?

Answer 1

The series

`sum_(n=0)^oo n3^nx^n`

is convergent for `x in (-1/3,1/3)`

Explanation:

Given the series:

`sum_(n=0)^oo n3^nx^n`

We can apply the ratio test to determine the interval of values of `x` for which the series is convergent.

We then evaluate:

`abs (a_(n+1)/a_n) = abs ( ( (n+1)3^(n+1)x^(n+1))/(n3^nx^n)) = 3((n+1)/n)absx`

So we have that:

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

The series is then absolutely convergent for `absx < 1/3` and divergent for `abs x > 1/3`

In the cases where `abs x = 1/3` the ration test is indecisive and we have to analyze in detail:

(i) `x = 1/3`

`sum_(n=0)^oo n3^nx^n = sum_(n=0)^oo n3^n(1/3)^n = sum_(n=0)^oo n = oo`

(iI) `x = -1/3`

`sum_(n=0)^oo n3^nx^n = sum_(n=0)^oo n3^n(-1/3)^n = sum_(n=0)^oo (-1)^n n`

Now if we can consider the partial sums of even order, we have:

`s_(2N) = sum_(n=0)^(2N) (-1)^n n = sum_(n=0)^N 2n-(2n-1) = N`

while for the partial sums of odd order:

`s_(2N+1) = s_(2N) -(2N+1) = N-2N-1 = -N-1`

so the series is irregular.

In conclusion the series is convergent for `x in (-1/3,1/3)`