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

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 I don't see how to proceed from here.

• Does $b_n$ converge to zero? Commented Jan 20 at 15:50
• @SungjinKim good point. I have amended this into the question. Commented Jan 21 at 14:21

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.

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 $$\alpha\mu(\alpha)\leq\int^\alpha_{\alpha/2} \mu(|f|>t)\,dt$$

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

1. $$\infty=d_f(0)=\lim_{\alpha\rightarrow0}d_f(\alpha)=\lim_{\alpha\rightarrow0}d_g(\alpha)=d_g(0)$$,
2. $$\max\big(d_g(\alpha),d_f(\alpha)\big)=0$$ 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 $$\frac{s}{2}d_g(\alpha) This implies that $$\|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}

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 $$\liminf_{\alpha\to 0^+} \frac{f(\alpha)}{g(\alpha)} = 0$$.