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

1 Answer
Jun 4, 2017

Radius of convergence: #1#

Explanation:

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#.