If $\{\lambda_n\} \subset \Bbb R_{>0}$, then $\limsup_m \left(\frac{\lambda_{n_m + 1}}{\lambda_{n_m}} \right) \le \liminf_n \, (\lambda_n)^{1/n}$

Question

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.

Answer 1

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

• Lim Sup and Lim Inf relations between the root test and the ratio test or
• Inequality involving $\limsup$ and $\liminf$: $ \liminf(a_{n+1}/a_n) \le \liminf((a_n)^{(1/n)}) \le \limsup((a_n)^{(1/n)}) \le \limsup(a_{n+1}/a_n)$)

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

Answer 2

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}