How do I compute the radii of convergence of the power series Σanx^n

(from n=0 to infinity)

with coefficients an=n and an=1/n

Printable View

- January 8th 2006, 09:54 AMTexasGirlRadii of convergence of a power series
How do I compute the radii of convergence of the power series Σanx^n

(from n=0 to infinity)

with coefficients an=n and an=1/n - January 8th 2006, 10:27 AMThePerfectHackerQuote:

Originally Posted by**TexasGirl**

Now for the second problem, use the ratio test for power series again to get Thus, the radius of converges is 1, thus for converges absolutely. Checking the endpoint (because the ratio test is inconclusive for when its limit is one), we have

but this is the alternating-harmonic series thus it converges. For the second possibility we have that but this is the harmonic series thus it diverges. Thus, the interval of convergence for the second power series is - January 9th 2006, 12:35 PMTexasGirlOne More Radius of Convergence Question...
Building from the same power series, if I take q, a nonzero element of C, and put it into the series so that I have the sum of an(q^n)(x^n), can I still use the ratio test in order to find the radius of convergence? Is an the coefficient?

- January 9th 2006, 01:54 PMThePerfectHackerQuote:

Originally Posted by**TexasGirl**

- January 9th 2006, 02:14 PMTexasGirl
Here is the question I have to answer:

Assume that the power series ∑an(x^n), where n=0 to infinity, has radius of convergence p, where p is a nonnegative real number or stands for the symbol ∞. Let q be an element of C, q≠0. Compute the radius of convergence of ∑an(q^n)(x^n) (n=0 to infinity). - January 10th 2006, 12:56 PMThePerfectHacker
As I understand it, its radius of convergence would be Explanation:

If has radius of convergence of then by the ratio test for power series because as I said it is the reciprocal of the limit. Thus, the new infinite series given by then by the ratio test thus, the radius of convergent is the reciprocal thus - January 11th 2006, 09:47 AMTexasGirlthanks again...
much appreciated... :)

- January 11th 2006, 02:21 PMThePerfectHacker
You are welcome.