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

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- Jan 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 - Jan 8th 2006, 10:27 AMThePerfectHackerQuote:

Originally Posted by**TexasGirl**

Now for the second problem, $\displaystyle a_k=1/k, k>1$ use the ratio test for power series again to get $\displaystyle \lim_{k\rightarrow \infty}|\frac{k}{k+1}|=1$ Thus, the radius of converges is 1, thus for $\displaystyle -1<x<1$ converges absolutely. Checking the endpoint (because the ratio test is inconclusive for when its limit is one), we have

$\displaystyle \sum^\infty_{k=1} (-1)^k 1/k$ but this is the alternating-harmonic series thus it converges. For the second possibility we have that $\displaystyle \sum^\infty_{k=1} 1/k$ but this is the harmonic series thus it diverges. Thus, the interval of convergence for the second power series is $\displaystyle -1\leq x<1$ - Jan 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?

- Jan 9th 2006, 01:54 PMThePerfectHackerQuote:

Originally Posted by**TexasGirl**

- Jan 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). - Jan 10th 2006, 12:56 PMThePerfectHacker
As I understand it, its radius of convergence would be $\displaystyle p/q$ Explanation:

If $\displaystyle \sum^{\infty}_{k=1}a_kx^k$ has radius of convergence of $\displaystyle p$ then by the ratio test for power series $\displaystyle \lim_{k\rightarrow \infty}|\frac{a_{k+1}}{a_k}|=1/p$ because as I said it is the reciprocal of the limit. Thus, the new infinite series given by $\displaystyle \sum^{\infty}_{k=1}a_kq^kx^k$ then by the ratio test $\displaystyle \lim_{k\rightarrow \infty}|\frac{a_{k+1}q^{k+1}}{a_kq^k}|=\frac{q}{p}$ thus, the radius of convergent is the reciprocal thus $\displaystyle \frac{p}{q}$ - Jan 11th 2006, 09:47 AMTexasGirlthanks again...
much appreciated... :)

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