How do you find the radius of convergence `sum_(n=1)^infty root(n)(n) x^n`?

Step 1. Perform the ratio test for absolute convergence, which says

If `lim_(n->infty)abs((a_(n+1))/(a_n))=L`

Then,

(i) If `L < 1`, the series is absolutely convergent
(ii) If `L > 1` or `lim_(n->infty)abs((a_(n+1))/(a_n))=infty`, then divergent
(iii) If `L=1`, apply a different test; the series may be absolutely convergent, conditionally convergent, or divergent.

So, applying the ratio test we get:

`lim_(n->infty)abs(((n+1)^(1/(n+1))x^(n+1))/(n^(1/n)x^n))=lim_(n->infty)abs(((n+1)^(1/(n+1))x)/(n^(1/n)))`

Since the denominator `n^(1/n)` doesn't approach zero as `n -> infty`, we can use the quotient rule.

`(lim_(n->infty)abs((n+1)^(1/(n+1)))x)/(lim_(n->infty)abs(n^(1/n)))`

Because the powers tend toward zero as `n->infty`, we can see that

`(lim_(n->infty)abs((n+1)^(1/(n+1))x))/(lim_(n->infty)abs(n^(1/n)))=abs(x xx 1)=abs(x)`

Step 2. Apply the constraint for absolute convergence.

IT follows from the ratio test that this series is absolutely convergent if `abs(x)<1`, that is, if `x` is in the open interval `(-1,1)`. The series diverges if `x > 1` or `x < -1`. Then numbers `1` and `-1` must be investigated separately by substitution in the power series.

Thus the interval of convergence is `-1 < x < 1` and the radius of convergence is the distance from the center point of the interval of convergence. So the radius of convergence is `1`.