# If $\sum b_n$ diverges, and $\sum a_n$ converges, and $b_n$ is monotonic does $\lim_{n\to\infty} \frac{a_n}{b_n}$ exist?

$$a_n,b_n>0$$

I've determined $$b_n$$ is monotonically increasing then $$\frac{1}{b_n}$$ is monotonically decreasing and bounded, so $$\sum \frac{a_n}{b_n}$$ converges and so the limit indeed exists and is equal to zero.

However, I'm not sure if $$b_n$$ is monotonically decreasing. Intuitively, $$a_n$$ must decrease faster than $$b_n$$, because the series converges, but I don't know how to prove that the limit must then exist.

• If $b_n$ is monotonic decreasing then $-b_n$ is monotonic increasing. Can you see how that helps? Commented Feb 18 at 0:21
• $b_n=-\frac 1 n$ is increasing but $\frac 1 {b_n}$ is not bounded. So your first sentence is wrong. Commented Feb 18 at 0:25
• My bad, I meant to add $b_n$ and $a_n$ are both greater than zero Commented Feb 18 at 0:26

The main flaw in the intuition here seems to be that any sequence must be either increasing or decreasing. There are many sequences that are neither, and one can construct many counterexamples this way.

Consider the following: define $$b_n=\frac1n$$ for all $$n$$. Define $$a_n = \begin{cases} \frac1n, &\text{if n is a power of 2}, \\ 0, &\text{otherwise}. \end{cases}$$ Then $$\sum b_n$$ diverges, but $$\sum a_n = \sum_{k=0}^\infty \frac1{2^k} = 2$$ converges. However, $$\frac{a_n}{b_n}$$ equals $$1$$ infinitely often and $$0$$ infinitely often, and thus $$\lim_{n\to\infty} \frac{a_n}{b_n}$$ does not exist. (If one requires that $$a_n$$ is strictly positive, one can just change the $$0$$s to incredibly small values—even $$\frac1{n^2}$$ is still enough to preserve the counterexample.)

• What happens if we require $a_n$ to be monotonic as well? Commented Feb 26 at 21:28

No, $$\lim_{i \rightarrow \infty} \frac{a_i}{b_i}$$ does not need to exist and this holds even with both $$a_i$$, $$b_i$$s required to be decreasing monotonically with $$i$$, and everything positive.

1. Let $$b_i = \frac{1}{i \log i}$$. Then the $$b_i$$s fit the necessary conditions. Indeed, $$\lim_{n \rightarrow \infty} \sum_{i=2}^n \frac{1}{i \log i} =\infty,$$ and the $$b_i$$s decrease with $$i$$.

2. Let us now define a sequence of integers $$n_k$$; $$k=1,2,\ldots$$ as follows: For each $$k=1,2,\ldots$$, let $$n_k := 2^{k^2}$$. Then note the following:

$$\sum_{k=1}^{\infty} n_kb_{n_k} = \sum_{k=1}^{\infty} 2^{k^2}\frac{1}{k^2 2^{k^2}}$$ $$=\sum_{k=1}^{\infty} \frac{1}{k^2} < \infty.$$

1. So, let $$a'_i$$ be the following sequence: Let $$k(i)$$ be defined to be the integer such that $$n_{k(i)-1} < i \le n_{k(i)}$$, or equivalently, $$2^{(k(i)-1)^2} < i \le 2^{k^2(i)}$$ for each positive integer $$k$$, is as defined above in 2. Then $$a'_i$$ is defined $$a'_i = b_{n_{k(i)}}$$ $$= \frac{1}{n_{k(i)} \log n_{k(i)}} = \frac{1}{k^2(i)2^{k^2(i)}}.$$ Then on the one hand: $$\sum_{i=1}^{\infty} a'_i \quad = \quad \sum_{k=1}^{\infty} \Big(\sum_{i=n_{k-1}}^{n_k} a'_i\Big)$$

$$= \sum_{k=1}^{\infty} (2^{k^2}-2^{(k-1)^2})\times \Big(\frac{1}{k^22^{k^2}}\Big)$$ $$\le \sum_{k=1}^{\infty} 2^{k^2} \times \Big(\frac{1}{k^22^{k^2}}\Big)$$

$$= \sum_{k=1}^{\infty} \frac{1}{k^2}$$ $$=\sum_{k=1}^{\infty} n_kb_k < \infty.$$ On the other hand: $$\frac{a'_{n_k}}{b_{n_k}} = 1$$ for all positive integers $$k$$.

1. Now, it is easy to get the desired $$a_i$$s from the $$a'_i$$s: Say set $$a_i := \frac{(k(i) \pmod 4)(1+e^{-e^{e^i}})}{k^2(i)2^{k^2(i)}}.$$

The idea is to perturb the $$a'_i$$s just a bit so that the resulting sequence $$a_i$$ is monotonically decreasing and $$\frac{a_{n_k}}{b_{n_k}}$$; $$k \rightarrow \infty$$ has no limit i.e., occollating around a few values, while maintaining $$\sum_{i=1}^{\infty} a_i$$ converging. Then, this does the trick.