Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
I can't imagine there being any simple, useful, and comprehensive answer to this question. Maybe you should experiment a little, find some examples and counterexamples, pose a hypothesis or two, try to prove a small result. – Gerry Myerson May 19 '11 at 0:29
I agree with you, Gerry. I thought I would ask the community and see if anyone has seen a similar result anywhere before. – wckronholm May 19 '11 at 0:49
For $\sum \frac{f(n)}{g(n)}$ to converge, $g(n)$ must grow significantly faster thatn $f(n)$ (or decrease much slower). However, given $f(x)$, we can perturb it just slightly to get arbitrary specified values at $f'(n)$, so for there to be any hope of the kind of relationship you want (which I doubt exists), you would need to restrict to a very nice class of functions (e.g. entire analytic). Since $\frac{f(x)}{g(x)}$ and $\frac{f'(x)}{g'(x)}$ have no relation except in indeterminate cases, I think you should follow Gerry's advice. – Aaron May 19 '11 at 0:51
Though I want to add that this does seem to work for polynomials, as in this case, convergence boils down to degree constraints. – Aaron May 19 '11 at 0:52
@Aaron: For most polynomials. If $f(n)=1$ and $g(n)=n$, then the statement I'm after fails for series. Of course, in this case, l'Hospitals rule fails to apply to the sequences themselves. – wckronholm May 19 '11 at 0:59
up vote 10 down vote accepted

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.

share|cite|improve this answer

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:

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

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

  3. $f$ and $g$ have "L'Hopital conditions": $\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow\infty}g(x)=0$ or $\pm\infty$.

share|cite|improve this answer

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....


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 $C^\infty$.

share|cite|improve this answer
You make some good points, but the issues you raise seem to disappear if we insist that f and g are monotonic. – wckronholm May 19 '11 at 4:52
Are you still able to demonstrate a counterexample when $f'$ and $g'$ are monotonic? I had similar counterexamples that led to the addition of this condition. Once f and g have concavity that is either always positive or always negative, I've lost my method for producing a counterexample. – alex.jordan May 21 '11 at 22:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.