Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have to proof this, but I don't know how to do this.

Let $R$ be the radius of convergence for $\sum_{k=0}^\infty a_k(x-a)^k$ and suppose that $\displaystyle\lim_{n \to +\infty} \left|\frac{a_{n+1}}{a_n}\right|=L$. Then:

(a) if $L$ is a nonzero finite real number, $R = \frac{1}{L}$,

(b) if $L=0, R = \infty$,

(c) if $L=\infty, R=0$.

share|improve this question
    
Hint: apply the ratio test. –  David Mitra Dec 12 '11 at 9:02
    
Which gives $(x-a) \frac{a_{k+1}}{a_k} \le (x-a) |\frac{a_{k+1}}{a_k}|$, but what to do next? –  Kevin Dec 12 '11 at 9:07
    
I gave an answer; but try to work it out from just the initial hint. :) –  David Mitra Dec 12 '11 at 9:17
add comment

1 Answer

The only thing you have to do is apply the Ratio test to the terms ${|a_n(x-a)^n|}$. Everything will drop out fairly easily by using the definition of the radius of convergence.



Warning: Solution follows:

Compute, for $x\ne a$: $$\lim_{n\rightarrow\infty}{|a_{n+1}| |x-a|^{n+1}\over |a_n||x-a|^n } =\lim_{n\rightarrow\infty}{|a_{n+1}||x-a| \over |a_n| } =|x-a| \lim_{n\rightarrow\infty}{|a_{n+1}| \over |a_n| }=|x-a|L. $$

The Ratio test allows you to conclude that the series converges whenever $|x-a|L<1$ and diverges whenever $|x-a|L>1$.

From the above, we can say:

If $L=0$, then the series converges for all $x$ and the radius of convergence is infinite.

If $L$ is infinite, then the series converges for no $x\ne a$. But the series does converge for $x=a$ (as trivially seen) and the radius of convergence is 0.

Otherwise, series converges whenever $|x-a| <{1\over L}$ and diverges whenever $|x-a| >{1\over L}$; which implies that the radius of convergence is $1\over L$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.