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This paper states Lemma $2.3$ without proof. I am trying to come up with a proof for the same.

Lemma $2.3$: If $\{\lambda_n\}_{n\in \mathbb N}$ is any sequence of positive numbers, then there exists a subsequence $\{n_m\}$ of $\mathbb N$ such that $$\limsup_m \left(\frac{\lambda_{n_m + 1}}{\lambda_{n_m}} \right) \le \liminf_n \, (\lambda_n)^{1/n}$$

The sequence $\{\lambda_n\}$ may or may not be convergent. Also, the lemma looks quite mysterious - how does the exponent $1/n$ show up on the right-hand side? I wonder if this follows from a more general result. Regardless, I'd appreciate any help with proving the statement above.

My thoughts: Suppose $\{\lambda_n\}$ is convergent, i.e., $\lambda_n \to \lambda \ge 0$. Consider the case $\lambda > 0$. Let $\{n_m\} = \mathbb N$, i.e., take the full sequence $\{\lambda_n\}$ as the subsequence. Then, $\limsup_n \frac{\lambda_{n+1}}{\lambda_n} = 1$ and $\liminf_n (\lambda_n)^{1/n} = 1$. So, equality holds. The cases left to consider are (i) $\lambda = 0$ and (ii) $\{\lambda_n\}$ is not convergent.

P.S. The authors claim that the result is well-known and elementary, but I've never seen it before.

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  • $\begingroup$ Maybe this is (partly) to do with the fact that, if you can prove convergence with the Root test, then you can prove convergence with the Ratio test. For example, see here: math.stackexchange.com/questions/287932/… $\endgroup$ Apr 4, 2023 at 8:39
  • $\begingroup$ It is a well known result from Calculus: For any sequence $(a_n)\subset X$, \begin{align} \liminf\frac{|a_{n+1}|}{|a_n|}\leq\liminf\sqrt[n]{|a_n|}\leq \limsup \sqrt[n]{|a_n|}\leq \limsup\frac{|a_{n+1}|}{|a_n|} \end{align} $\endgroup$
    – Mittens
    Apr 6, 2023 at 11:12
  • $\begingroup$ What's $X$? A normed vector space? A proof, a reference, or at least some ideas would be helpful. Thank you. @OliverDíaz $\endgroup$ Apr 6, 2023 at 11:14
  • $\begingroup$ @esoteric-elliptic: $X$ should have been $\mathbb{C}$, but the whole thing works for $X$ a normed space. $\endgroup$
    – Mittens
    Apr 6, 2023 at 11:16

2 Answers 2

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It is well-known that $$ \liminf_{n \to \infty} \left(\frac{\lambda_{n + 1}}{\lambda_{n}} \right) \le \liminf_{n \to \infty} \, (\lambda_n)^{1/n} \, . $$ holds for all sequences $(\lambda_n)$ of positive real numbers, see for example

It only remains to choose a subsequence $(n_m)$ of $\Bbb N$ such that $$ \lim_{m\to \infty} \left(\frac{\lambda_{n_m + 1}}{\lambda_{n_m}} \right) = \liminf_{n \to \infty} \left(\frac{\lambda_{n + 1}}{\lambda_{n}} \right) \, . $$

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It is a result typically seen in Calculus courses in the study of convergence of series (ratio and root test) and also in Complex variable courses (radius of convergence of Power series).

Lemma: For any sequence $(a_n)\subset \mathbb{C}\setminus\{0\}$, \begin{align} \liminf\frac{|a_{n+1}|}{|a_n|}\leq\liminf\sqrt[n]{|a_n|}\leq \limsup \sqrt[n]{|a_n|}\leq \limsup\frac{|a_{n+1}|}{|a_n|} \end{align}

Proof: Denote by $\beta^*=\limsup_n\frac{|a_{n+1}|}{|a_n|}$, $\alpha^*=\limsup \sqrt[n]{|a_n|}$, $\beta_*=\liminf\frac{|a_{n+1}|}{|a_n|}$ and $\alpha_*=\liminf \sqrt[n]{|a_n|}$. If $\beta^*<\infty$, then for any $b>\beta^*$ fixed, there exists $N\in\mathbb{N}$ such that $$ |a_{n+1}|<b|a_n|\qquad\text{for all}\qquad n\geq N. $$ It follows that $|a_{m+N}|\leq b^m |a_N|$ for all $m>0$; consequently, $$ \sqrt[n]{|a_n|}\leq b^{1-N/n}\sqrt[n]{|a_N|} $$ for $n>N$. Letting $n\rightarrow\infty$ and then $b\searrow\beta^*$ shows that $\alpha^*\leq\beta^*$. A similar argument shows that $\beta_*\leq \alpha_*$.

Finally, choose a subsequences $m_k$ and $n_k$ (which exists from definition of $\limsup$ and $\liminf$) such that \begin{align} \beta_*=\lim_k\frac{|a_{m_k+1}|}{|a_{m_k}|},\qquad \beta^*=\lim_k\frac{|a_{n_k+1}|}{|a_{n_k}|} \end{align} to get `\begin{align} \limsup_k\frac{|a_{m_k+1}|}{|a_{m_k}|}&= \lim_k\frac{|a_{m_k+1}|}{|a_{m_k}|}\leq \liminf\sqrt[n]{|a_n|}\\ &\leq\limsup\sqrt[n]{|a_n|}\leq \lim_k\frac{|a_{n_k+1}|}{|a_{n_k}|}=\liminf_k\frac{|a_{n_k+1}|}{|a_{n_k}|} \end{align}

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  • $\begingroup$ This doesn't seem to prove the inequality required in the post, which is (i) along a subsequence and (ii) basically the opposite inequality. Could you clarify? $\endgroup$ Apr 6, 2023 at 12:10
  • $\begingroup$ Alright, I'll try it out and get back if any questions arise! Thanks again. $\endgroup$ Apr 6, 2023 at 13:20

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