0
$\begingroup$

I found a definition for the radius of convergence of a power series: $$r =\ \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}}\right|$$

or $$r =\ \lim_{n \to \infty}\frac{1}{\sqrt[n]{|a_n|}}$$ if the limit exists .

I know that this comes from the ratio and root test , and I solved many problems about radius of convergence..

But I would like to know what is the meaning of " the limit exists " here.

Does it mean that if the limit is infinity then there is NO radius of convergence , because it will converge for all values of x ? (However , my teacher said that if limit is infinity then the radius of convergence exists and is equal to infinity .

OR it means that in some cases , we will find the limit has more than one value or we can not find its value ( like if we take the limit for sin(n) )

$\endgroup$
  • $\begingroup$ A power series has always a radius of convergence $R\in [0,+\infty] $. $\endgroup$ – hamam_Abdallah Dec 19 '17 at 19:27
  • $\begingroup$ So the " limit does not exist " here just means that we can not find a specific value for the limit ( like if we take the limit for sin(n) ) ? $\endgroup$ – MCS Dec 19 '17 at 19:33
  • $\begingroup$ Even if the limit does not exist, the radius exists. $\endgroup$ – hamam_Abdallah Dec 19 '17 at 19:37
  • $\begingroup$ The purpose of these tests is to make a comparison with a geometric series. math.stackexchange.com/questions/2202516/…. $\endgroup$ – zwim Dec 19 '17 at 19:38
  • 1
    $\begingroup$ According to Terence Tao, think in limsup and liminf, which always exists. (can be finite or infinity) "Limit exists" can thus be translated to "liminf=limsup". $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 19 '17 at 19:52
1
$\begingroup$

"The limit $ \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}}\right| $ exists" in this context means

  • the sequence of real number $\{| \frac{a_{n}}{a_{n+1}}|\}$ is convergent: there exists a real number $L$ such that $$ \forall\epsilon>0\ \exists N\in{\bf N}\ \forall n\geq N\ \ |\frac{a_{n}}{a_{n+1}}-L|<\epsilon $$
  • or $| \frac{a_{n}}{a_{n+1}}|\to\infty$ as $n\to\infty$.

(The other one this similar.) In the second case, it is a convention to say that the raidus of convergence is $\infty$, although $\infty$ is not a real number.

Given a sequence of complex numbers $\{a_n\}$, the two limits $$ \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}}\right|, \quad \lim_{n \to \infty}\frac{1}{\sqrt[n]{|a_n|}} $$ may or may not exist. If one of them (or both) exists, then it is defined as the the radius of convergence of the corresponding power series.

If they both don't exist, then the definition says nothing, but we still have the concept of "radius of convergence": see the Cauchy–Hadamard theorem.

$\endgroup$

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.