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

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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?


1 Answer 1


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.

  • $\begingroup$ "personally, I think it makes no sense to write it like that" Sorry but what exactly "makes no sense" here? $\endgroup$
    – Did
    Jan 24, 2018 at 19:43
  • $\begingroup$ Then, how can I calculate $\lim 2^{\frac{-n}{2n-1}}$? $\endgroup$
    – Abdu Magdy
    Jan 24, 2018 at 20:10
  • $\begingroup$ @Did: writing the liminf as the limit of a sequence that is not a subsequence. $\endgroup$ Jan 24, 2018 at 20:47
  • 1
    $\begingroup$ @AbduMagdy: I have edited that into the answer. $\endgroup$ Jan 24, 2018 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, 2018 at 21:09

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