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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?

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Just subtracting one series from the other, it seems that you need $\sum_{n = 0}^{\infty} (f(n)-g(n))$ to converge. (This is what fails in Xandi Tuni's example.) Writing $f(n)/g(n) = 1 + \delta_n,$ so that $\sum_{n = 0}^{\infty} (f(n) - g(n)) = \sum_{n = 0}^{\infty} \delta_n g(n),$ you see need control over the signs of the $\delta_n$. 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 are in bad shape. (This is what goes wrong in Xandi Tuni's example.) –  Emerton May 13 '10 at 14:15
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Emerton's comment basically does it, I think --- you need to be able to control either the signs of the $\delta_n$s or their sizes --- if $\sum \delta_n$ converges absolutely, for example, then you're golden, and this requires only that the approximation $f(n) \approx g(n)$ gets sufficiently better as $n\to \infty$. Emerton: you should leave your comment as an answer. –  Theo Johnson-Freyd May 13 '10 at 17:52
    
I was wondering why wood's second question was also numbered one, so I went to edit the file and...wow! If you like jsmath mysteries, look into this one. (Note that I tried putting gray boxes around the text to solve the problem. It didn't solve it, but after I hacked my way out I decided the gray boxes looked nice and left them there.) –  Pete L. Clark May 13 '10 at 22:44
    
@Pete, that's markdown trying to being too smart for its own good. It insists in taking care of numbered list, and its mind is too litle to see that one may want to intersperse extra text between items in the list... Escaping the numbers with backticks fixes the numbering but loses the formatting :/ –  Mariano Suárez-Alvarez May 13 '10 at 23:07
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3 Answers

up vote 2 down vote accepted

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.)

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I think you want $f(n)/g(n)=1+\delta_n$. –  Kevin Ventullo May 25 '10 at 3:53
    
Thanks Kevin; this is now fixed. –  Emerton May 25 '10 at 18:35
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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.

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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.

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