Non-absolute convergence of series with asymtotically equal coefficients - MathOverflow

Non-absolute convergence of series with asymtotically equal coefficients

2010-05-13T12:41:43Z

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 by Xandi Tuni for Non-absolute convergence of series with asymtotically equal coefficients

2010-05-13T13:10:04Z

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 by Julián Aguirre for Non-absolute convergence of series with asymtotically equal coefficients

2010-05-13T19:28:52Z

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.

Answer by Emerton for Non-absolute convergence of series with asymtotically equal coefficients

2010-05-14T07:57:19Z

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.)