Suppose $(a_n),(b_n)$ are decreasing sequences, $\sum a_n$ converges, $\sum b_n$ diverges. $\liminf_{\alpha\to 0^+}\frac{n_a(\alpha)}{n_b(\alpha)}=0?$

Question

Suppose $(a_n), (b_n)$ are positive, decreasing real sequences, $\displaystyle\sum a_n$ converges, $\displaystyle\lim_{n\to\infty} b_n=0,$ and $ \displaystyle\sum b_n$ diverges. For each $\alpha > 0,$ define $f(\alpha)$ to be the least integer $n$ such that $a_n< \alpha.$ Similarly, define $g(\alpha)$ to be the least integer $m$ such that $b_m< \alpha.$ Prove or disprove: $\displaystyle\liminf_{\alpha\to 0^+} \frac{f(\alpha)}{g(\alpha)} = 0.$

This question arose when thinking about this question, where the counter-examples in the answers are clearly also counter-examples to the proposition, "$\displaystyle\lim_{\alpha\to 0^+} \frac{f(\alpha)}{g(\alpha)} = 0.$"

I can't think of any counter-examples to the above question though. With the examples in the question linked above, I think $\displaystyle\liminf_{\alpha\to 0^+} \frac{f(\alpha)}{g(\alpha)} = 0.$

I guess the standard approach to prove the above question true is proof by contradiction. So suppose $\displaystyle\liminf_{\alpha\to 0^+} \frac{f(\alpha)}{g(\alpha)} \neq 0,\implies \displaystyle\liminf_{\alpha\to 0^+} \frac{f(\alpha)}{g(\alpha)} > 0,\ $ that is, $\ \exists\ \varepsilon>0,\ \exists\ t>0,\ $ such that $\frac{f(t')}{g(t')} > \varepsilon\ \forall\ 0<t'<t.\ $ I don't see how to proceed from here.

Answer 1

If $\liminf_{\alpha\to 0^+} \frac{f(\alpha)}{g(\alpha)} > 0$ then there is an $\epsilon > 0$ and a $\alpha_0 > 0$ such that $\frac{f(\alpha)}{g(\alpha)} \ge \epsilon$ for $0 < \alpha \le \alpha_0$. The idea is to use summation by parts to show that $\sum b_n$ can be estimated above in terms of $\sum a_n$, and in particular that $\sum b_n < \infty$ in contradiction to the assumption that $\sum b_n$ diverges.

Remark: I'll assume that the sequence indices of $(a_n)$ and $(b_n)$ start at $n=1$ and use $\Bbb N$ for the set of positive integers $1, 2, 3, \ldots$.

Let $(x_k)_{k \ge 1}$ be the distinct element of $$ \{ a_n \mid n \in \Bbb N \} \cup \{ b_n \mid n \in \Bbb N \} $$ in decreasing order and choose some $x_0 > x_1$. Note that $x_k \to 0$.

Let $0 < \delta < \alpha_0$. Let $M$ be the first index with $x_{M} \le \alpha_0$, and let $m$ be the first index with $x_{m+1} \le \delta$. Then $$ \begin{align} \sum_{b_n > \delta} b_n &= \sum_{k=1}^m \bigl(g(x_k) - g(x_{k-1})\bigr) x_k\\ &= \sum_{k=1}^m g(x_k) x_k - x_{k-1} g(x_{k-1}) + g(x_{k-1}) (x_{k-1} - x_k) \\ &= -x_0g(x_0) + \left(\sum_{k=1}^m g(x_{k-1}) (x_{k-1} - x_k) \right) + g(x_m)x_m \\ &= C_1 + \left(\sum_{k=M+1}^m g(x_{k-1}) (x_{k-1} - x_k) \right) + g(x_m)x_m \end{align} $$ for some constant $C_1$ which does not depend on $\delta$. In the same way we get $$ \sum_{a_n > \delta} a_n = C_2 + \left(\sum_{k=M+1}^m f(x_{k-1}) (x_{k-1} - x_k) \right) + f(x_m)x_m $$ for some constant $C_2$ which does not depend on $\delta$. Now $g(x_k) \le \frac 1 \epsilon f(x_k)$ for $k \ge M$, so that $$ \sum_{b_n > \delta} b_n \le C_1 - \frac{C_2}{\epsilon} + \frac 1 \epsilon \sum_{a_n > \delta} a_n \, . $$ This holds for all $\delta > 0$, and we get $$ \sum_{n=1}^\infty b_n \le C_1 - \frac{C_2}{\epsilon} +\frac 1 \epsilon \sum_{n=1}^\infty a_n < \infty $$ in contradiction to the assumptions. This proves that $ \liminf_{\alpha\to 0^+} \frac{f(\alpha)}{g(\alpha)} = 0 $.

Answer 2

Since convergence (or divergence) and summation of series of absolutely convergent series is invariant under rearrangements, I only consider the case where the series $a$ and $b$ are positive and non increasing in this posting.

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Some general facts:

Consider the space $(\mathbb{N},2^{\mathbb{N}},\mu)$ where $\mu$ is the counting measure. Let $\mathcal{c}_0$ denote the space of bounded sequences $a$ such that $\lim_{n\rightarrow\infty}a(n)=0$. Notice that the space of absolute convergent series is $L_1(\mu)$ with norm $$\|f\|_1=\int_\mathbb{N}|f(n)|\mu(dn)=\sum^\infty_{n=1}|f(n)|$$ For any sequence $f$, define $$d_f(\alpha)=\mu(|f|>\alpha),\qquad \alpha\geq0$$ Notice that $d_f$ is a right-continuous, monotone non increasing and that $\lim_{\alpha\rightarrow\infty}d_f(\alpha)=0$ if $d_f(s)<\infty$ for some $s\geq0$.

If $f\in\mathcal{c}_0$ is positive ($f(n)>0$ for all $n\in\mathbb{N})$ and non increasing, then \begin{align} n_f(\alpha):=\min\{n\in\mathbb{N}: f(n)<\alpha\}= d_f(\alpha-)+1, \quad 0\leq \alpha<\|f\|_\infty \tag{0}\label{zero} \end{align}

It is known that for $f\in L_1(\mu)$,

1. $\|f\|_1=\int^\infty_0d_f(t)\,dt$,
2. and $\lim_{\alpha\rightarrow0+}\alpha d_f(\alpha)=\lim_{\alpha\rightarrow\infty}\alpha d_f(\alpha)=0$ for $$\alpha\mu(|f|>\alpha)\leq\int_{|f|>\alpha}|f|\leq\|f\|_1,$$ the fact that $\|f\|_1=\int^\infty_0 d_f(\alpha)\,dt$, and $$\alpha\mu(\alpha)\leq\int^\alpha_{\alpha/2} \mu(|f|>t)\,dt$$

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Application to the OP's problem:

Suppose $f,g$ are positive (not necessarily monotone) sequences such that $f\in L_1(\mu)$, $g\notin L_1(\mu)$, and such that $\lim_{n\rightarrow\infty}g(n)=0$. Then

3. $\infty=d_f(0)=\lim_{\alpha\rightarrow0}d_f(\alpha)=\lim_{\alpha\rightarrow0}d_g(\alpha)=d_g(0)$,
4. $\max\big(d_g(\alpha),d_f(\alpha)\big)=0$ for all $\alpha>\max(\|f\|_\infty,\|g\|_\infty)$.

• If $s:=\liminf_{\alpha\rightarrow0}\frac{d_f(\alpha)}{d_g(\alpha)}>0$, then there is $\alpha_0>0$ such that $$\frac{s}{2}d_g(\alpha) <d_f(\alpha), \qquad 0<\alpha<\alpha_0$$ This implies that $$\|g\|_1=\int^\infty_0 d_g(\alpha)\,d\alpha<\infty$$ in contradiction to $\|g\|_1=\infty$. Therefore, \begin{align} \liminf_{\alpha\rightarrow0}\frac{d_f(\alpha)}{d_g(\alpha)}=0\tag{5}\label{five} \end{align}

• If in addition $f$ and $g$ are also nonincreasing, then by \eqref{zero}, observation [3], and \eqref{five} \begin{align} \liminf_{\alpha\rightarrow0}\frac{n_f(\alpha)}{n_g(\alpha)}=0\tag{6}\label{six} \end{align}