$\lim \sup$ and $\lim \inf$ of root and ratio test sequences of a series (Rudin)

enter image description here enter image description here

I understand that:

$$(1) \qquad \{\frac{a_{n+1}}{a_n}\} = \{\frac{2}{3},\frac{3}{2^2},\frac{2^2}{3^2},\frac{3^2}{2^3},...\} = \{\frac{2^n}{3^n}, \frac{3^n}{2^{n+1}}\}$$

$$(2) \qquad \{\sqrt[n]{a_n}\} = \{\sqrt[1]{1/2},\sqrt[2]{1/3},\sqrt[3]{1/2^2},\sqrt[4]{1/3^2},...\} = \{\sqrt[2n-1]{\frac{1}{2^n}}, \sqrt[2n]{\frac{1}{3^n}}\}$$

Then there are two subsequences of $(1)$ and $(2)$, each of which will lead to $\lim \sup$ or $\lim \inf$ for each sequence.

1- Is my understanding right?
2- Why $\lim \sup \ (2)=\lim \sqrt[2n]{\frac{1}{2^n}}$ while $\{\sqrt[2n]{\frac{1}{2^n}}\}$ is not a subsequence of $(2)$?
3- Does $(1)$ or $(2)$ has only two subsequences?
4- Does $(1)$ or $(2)$ has only two subsequential limits?
5- Is there an informal way to choose the subsequence whose limit is $\lim \sup/\inf$ of a sequence, or should I consider every possible subsequence?


Your understanding is right.

The author just decided to write the limit in a way he likes, but it's not coming directly from the sequence; personally, I think it makes no sense to write it like that. To calculate the limit of the subsequence, $$ (a_{2n-1})^{1/(2n-1)}=(2^{-n})^{1/(2n-1)}=e^{-\frac{n}{2n-1}\,\log 2}\to e^{-\frac12\,\log 2}=2^{-1/2}=\frac1{\sqrt2}. $$

There are only two subsequential limits in this case, because the distance between $2^{-n/(2n-1)}$ and $3^{-1/2}$ cannot be small for large $n$.

There is no canonical recipe in general. These example works nicely because there are only two subsequences to choose from.

| cite | improve this answer | |
  • $\begingroup$ "personally, I think it makes no sense to write it like that" Sorry but what exactly "makes no sense" here? $\endgroup$ – Did Jan 24 '18 at 19:43
  • $\begingroup$ Then, how can I calculate $\lim 2^{\frac{-n}{2n-1}}$? $\endgroup$ – Abdu Magdy Jan 24 '18 at 20:10
  • $\begingroup$ @Did: writing the liminf as the limit of a sequence that is not a subsequence. $\endgroup$ – Martin Argerami Jan 24 '18 at 20:47
  • 1
    $\begingroup$ @AbduMagdy: I have edited that into the answer. $\endgroup$ – Martin Argerami Jan 24 '18 at 20:49
  • $\begingroup$ Well, $a_{2n-1}=1/2^n$ and $a_{2n}=1/3^n$ hence $a_{2n-1}^{1/(2n-1)}=(1/2^n)^{1/(2n-1)}$ and $a_{2n}^{1/2n}=1/3^{1/2}$. Are you complaining because R. replaced $(1/2^n)^{1/(2n-1)}$ by $(1/2^n)^{1/(2n)}$? Saying that "it makes no sense" is slightly too strong (hence, misleading), if you ask me. $\endgroup$ – Did Jan 24 '18 at 21:09

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.