When does l'Hospital's rule work for series? This question stems from a response a colleague of mine received as a (flawed) solution to a problem on his calculus exam. The student was to determine convergence of a series of the form $\displaystyle\sum_{n=1}^\infty \frac{f(n)}{g(n)}$. The student instead considered the series  $\displaystyle\sum_{n=1}^\infty \frac{f'(n)}{g'(n)}$ and thought that this series and the original must behave the same way. This, of course, is false in general.

What conditions on $f$ and $g$ are required so that the series  $\displaystyle\sum_{n=1}^\infty \frac{f(n)}{g(n)}$ converges iff  $\displaystyle\sum_{n=1}^\infty \frac{f'(n)}{g'(n)}$ converges?

 A: I'm the guy who had the confused student.
In my original posing of the question, I ask for $f'$ and $g'$ to be monotonic, not just $f$ and $g$.  Are you still able to create a counterexample?  
Someone mentioned usefulness.  If something like this worked, I'd have students apply it to series like n/(n^3-1).  The Comparison Test is awkward to use correctly.  The Limit Comparison Test is great for this one, but it's nice to have options.
Maybe the list of conditions to make this conjecture true becomes too lengthy to bother with.  But I have yet to see a counterexample to these conditions:


*

*$f'$ and $g'$ are monotonic.

*$\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}=0$

*$f$ and $g$ have "L'Hopital conditions": $\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow\infty}g(x)=0$ or $\pm\infty$.
A: In my oppinion any such result would be artificial and make very little mathematical sense (excepting of course for students).
How would we use such a result? Given the series $\displaystyle\sum_{n=1}^\infty \frac{f(n)}{g(n)}$, since $f,g$ are defined only for integers, what does $f'(n)$ and $g'(n)$ mean?
Backwards same problem, how do we integrate a function defined only on positive integers.
Of course, most of the exercises the students meet involve only elementary functions, so for students is pretty natural to derivate/integrate such sequences; yet in reality any function defined on the integers has uncountably many differentiable extensions to the real numbers.
Any such test would either need to work for all of them (I highly doubt it), or to excluse most of them.
Last but not least, keep in mind given any function of the type $\frac{f(n)}{g(n)}$ defined on the integers, it is easy to extend it (probably in countably many ways) to differentiable functions $f,g$ on ${\mathbb R}$ so taht $f'(n)=0$ and $g'(n)\neq 0$....
And this can be done most of the times so that basic properties of $f,g$ and $\frac{f}{g}$ such as monotony are presearved....
Just to include a simple example to emphasise the problem: 
Look to the series $\sum \frac{1}{n^2}$.
What would the derivative series be be? Obvious right $\sum \frac{0}{2n}$...
Or since our series is also the same as  $\sum \frac{1+\sin(2n \pi)}{n^2}$ then the derivative series should be $\sum \frac{2 \pi \cos(2n \pi)}{2n}=\sum \frac{ \pi }{n}$.
These two derivatives have completely different behavior... And if you think this is easy, if $h(x)$ is any differentiable function then our originar series is also:
$$\sum \frac{1+\sin(2n \pi)h(n)}{n^2} \,.$$
and now things really get messy....
Added
Asking for $f$, $g$ to be monotonic doesn't change too much, is just that counterexamples are a little harder to build. Lets say that $f$ is decreasing.
Fix an $n$. Then the function $h: [n,n+1]$ defined by 
$$h(x)= \frac{f(n)+f(n+1)}{2}+ \frac{f(n)-f(n+1)}{2}\cos((x-n)\pi) \,.$$
is decreasing, and $h(n)=f(n), h(n+1)=f(n+1)$ wile $h'(n)=h'(n+1)=0$. 
Doing this over each interval, you can replace any decreasing $f$ by a new function, which is monotonic and whose derivative is identically zero on the integers... 
And more complicated examples can be constructed so the derivative is any negative sequence, and I am pretty sure this can be done to be even $C^\infty$.   
A: Suppose $f$ and $g$ are meromorphic in a neighbourhood $U$ of $\infty$ with poles at $\infty$ (so $\lim_{z \to \infty} f(z) = \lim_{z \to \infty} g(z) = \infty$), having orders $p$ and $q$ respectively.  Then $f'$ and $g'$ have order $p-1$ and $q-1$ respectively at $\infty$.
If $\{N, N+1, \ldots\} \subset U$,
then $\sum_{n=N}^\infty \frac{f(n)}{g(n)}$ converges iff $q \ge p + 2$
iff $\sum_{n=N}^\infty \frac{f'(n)}{g'(n)}$ converges. 
