1
$\begingroup$

For a given power series $$S = \sum_{n}c_n (x - a)^n$$ We have interval of convergence for it defined as the set of values of $x$ for which the series converges.

Many reference books says I can use ratio test or root test for the radius, e.g. $$\limsup_{n\rightarrow+\infty}\sqrt[n]{\left | c_n (x - a)^n\right |} = \limsup_{n\rightarrow+\infty}\sqrt[n]{\left | c_n \right |}\left | x - a\right |< 1$$

Root test is a sufficient condition for convergence but never a necessary condition. If we use root test to find the interval of convergence, we'll arrive at a subset of the true interval.

And books give that we should check the endpoints of the interval. I could understand that if the series diverges on both endpoints, then the values outside the interval just diverges as well. But what if the series does converge on either endpoint or both? How do we confirm that there are no possible value outside the interval that still makes the series converge?

$\endgroup$
  • $\begingroup$ So you answered your own question? The tests can just say that you get convergence if $|x-a|<R$ and divergence if $|x-a|>R$. Since there are examples where some of $|x-a|=R$ converge and some not, these points have to be considered separately. But your test should consider $\limsup$ instead of $\lim$ since $(c_n)_n$ has not to be convergent. $\endgroup$ – Mundron Schmidt Jun 27 '17 at 7:47
  • $\begingroup$ @MundronSchmidt Thank you. Exactly the same idea, just not that obvious to me... $\endgroup$ – YiFei Jun 27 '17 at 7:55
  • $\begingroup$ And you should write $\sqrt[n]{|c_n|}$ instead of $\sqrt[n]{c_n}$ if $c_n$ is not positive or complex. $\endgroup$ – Mundron Schmidt Jun 27 '17 at 7:56
  • $\begingroup$ @MundronSchmidt Was being lazy... ;) $\endgroup$ – YiFei Jun 27 '17 at 7:59
0
$\begingroup$

Finally, I think I get the answer. (The key part is that root and ratio test are for both convergence and divergence.)

The root or ratio test in this process actually gives three part of results.

For some $x$, the limit is greater than 1, and the series diverges.

For some other $x$, the limit is less than 1, the series converges.

For the rest $x$, the result is unknown, and is left as endpoint. (Provided that the function in the limit is continuous)

So that only endpoint are left in undefined state that requires further work and which is tested with other methods.

$\endgroup$
  • $\begingroup$ Yes, you got it. $\endgroup$ – Mundron Schmidt Jun 27 '17 at 7:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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