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I teach freshman calculus, and have recently been discussing series. In one question on a recent test, I asked whether $\sum\frac{n^2}{n^3+1}$ converges or diverges.

One student got the correct answer by the following incorrect reasoning. They used L'Hopital's rule to conclude that $\frac{n^2}{n^3+1}$ has the same limit as $\frac{2n}{3n^2}$ and as $\frac{2}{6n}$ (which is true so far). They then claimed that because $\sum\frac{2}{6n}$ diverges, the original series must diverge.

This, of course, does not follow from anything they've been taught. L'Hopital's rule merely says that if $\frac{2}{6n}$ approaches zero, we can conclude that $\frac{n^2}{n^3+1}$ approaches zero. But L'Hopital's rule, as typically stated, says nothing about the "rate" at which $\frac{n^2}{n^3+1}$ approaches zero.

However, the following conjecture seems to be true for many expressions, such as $\frac{x^a}{e^x}$, $\frac{\ln x}{x^a}$, and $\frac{x}{x^2(\ln x)^a}$. Can anyone help me come up with the "correct" conjecture, and a proof?

CONJECTURE: If $f(x)$ and $g(x)$ belong to a "nice" class of functions (which approach infinity with $x$), then the infinite series $\sum_n\frac{f(n)}{g(n)}$ and $\sum_n\frac{f'(n)}{g'(n)}$ either both converge or both diverge.

EDIT: It appears essentially the same question has already been asked. And a nice answer was given by Robert Israel. However, he requires that $f$ and $g$ be meromorphic. It would be nice if there were a slightly more general answer that included logarithms, OR a counterexample using functions built out of logarithms.

When does l'Hospital's rule work for series?

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You're only interested in what happens for large arguments, and out there logarithms are analytic, no? –  Gerry Myerson Oct 29 '11 at 22:22
    
I think maybe I don't fully understand how the logarithm function looks when considered as a function of a complex variable, in a neighborhood of infinity. I don't remember much about the complex logarithm, and my ability to visualize it is hindered by the fact that it's multi-valued. –  idmercer Oct 29 '11 at 23:57
    
$\log z$ is analytic in what's left of the complex plane after you remove the non-positive real numbers. If $z=re^{i\theta}$, then $\log z=\log r+i\theta$, so it goes to infinity as $z$ does. I'm using the principal branch of the logarithm here, so it's not multi-valued, and $\theta$ is (strictly) between $-\pi$ and $\pi$. –  Gerry Myerson Oct 30 '11 at 2:05
2  
I see the problem: $\log z$ is not defined in a neighborhood of infinity, and I guess that's why Robert Israel's argument doesn't go through, and why Will Jagy's counterexample can exist. –  Gerry Myerson Oct 30 '11 at 2:11

1 Answer 1

up vote 7 down vote accepted

Take $f(x) = \log \log x,$ then $g(x) = x \log x.$ The sum of $1/g$ diverges, so the sum of $f/g$ also diverges. But $f' / g'$ is slightly smaller than $$ \frac{1}{x (\log x)^2} $$ and this sum converges.

For this, you need to notice that an antiderivative of $\frac{1}{x \log x}$ is $\log \log x,$ while an antiderivative of $\frac{1}{x (\log x)^2}$ is $\frac{-1}{\log x}.$

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Neat example. Thanks! –  idmercer Oct 30 '11 at 13:52

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