Non-absolute convergence of series with asymtotically equal coefficients

Question

The following seems to be a question related to standard calculus, but I am not quite sure where to look for an answer.

Suppose $f,g:\mathbb{N} \to \mathbb{C}$ are such that the have the same asymptotical behaviour, i.e. $f(n)/g(n) \to 1$ as $n \to \infty$. Of course, suppose that one of the sums $\sum_{n=0}^\infty f(n)$ and $\sum_{n=0}^\infty g(n)$ converges absolutely, then so does the other. This can be proven by a standard estimate. However this standard estimate fails if we do not have absolute convergence. I do not see how to prove convergence of one of sums implies the convergence of the other. I feel that it may be actually false.

So the first question is:

$1$. Is it true that one series converges iff the other does?

If this is not the case, however, in the problem I am studying, I want to prove convergence for both series. For my application in mind, you may assume that $f(n)/g(n)$ is always in $\mathbb{R}$. So the second question is

$2$. Under which additional conditions (which do not! imply absolute convergence) can we deduce both series have the same behaviour. Are there books treating such topics?

Answer 1

Following Theo Johnson--Freyd's suggestion, I am making my above comment an answer:

Just subtracting one series from the other, it seems that you need $\sum_{n=0}^{\infty} (f(n)−g(n))$ to converge. Writing $$f(n)/g(n)=1+\delta_n,$$ $$\text{so that } \qquad \qquad\sum_{n=0}^{\infty} (f(n)−g(n))=\sum_{n=0}^{\infty} \delta_n g(n),$$ you see need control over the signs of the $\delta_n$, or (as Theo notes in his comment, on their rate of growth).

E.g. if they are all of the same sign, you are okay, while if the sign of $\delta_n$ is always the same as, or always opposite to, that of $\delta_n$, then you could be in bad shape. (This is what goes wrong in Xandi Tuni's example.)

As Theo notes in his comment, you are also okay if $\sum_{n = 0}^{\infty} \delta_n$ converges absolutely. Whether this applies in your case will depend on how closely $f$ and $g$ approximate one another. (This is illustrated by the example in Julian Aguirre's answer.)

Answer 2

What about this: Take $f(n) = (-1)^n\log(n)^{-1}$ and $g(n)=f(n)+n^{-1}$. The sum of the $f(n)$'s is converging (it is "telescopic") and very far from absloutely converging. The sum of $g(n)$'s diverges to $+\infty$. Finally we have

$$f(n)/g(n) = \frac{1}{1 + (-1)^n\log(n)n^{-1}}$$

which goes to 1 as $n$ goes to infinity. So that means "no" for your first question.

Answer 3

This is a different example than the one given by Xandi Tuni. Let $a_n$ be any nondecreasing sequence of real numbers such that $a_1>1$ and $\lim_{n\to\infty}a_n=\infty$, and let $f(n)=(-1)^n/a_n$. Then $\sum f(n)$ converges by Leibniz's criterion. Now define $$ g(n)=\frac{(-1)^n}{a_n+(-1)^n} \implies \lim_{n\to\infty}\frac{f(n)}{g(n)}=\lim_{n\to\infty}\frac{a_n+(-1)^n}{a_n}=1. $$ Then $$ g(n)=f(n)-\frac{1}{a_n(a_n+(-1)^n)}, $$

so that $\sum g(n)$ converges if and only if $\sum\frac{1}{a_n^2}$ does.