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Are there rules for answering this question:

$f(x)$ is a smooth function that's always increasing without bound, and $g(x)$ is a smooth function that always decreases from 1, approaching zero but never equaling it. $\sum_{x=1}^nf(x)$ diverges as n approaches infinity. But we know for sure that $\sum_{x=1}^ng(x)$ converges as n approaches infinity. Will $\sum_{x=1}^nf(x)\cdot g(x)$ converge or diverge? $f(x)$ and $g(x)$ are positive for all $x\ge 1$.

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For the sums, $f$ and $g$ only need to be defined at the integers, in which case what do we make of derivatives? – Gerry Myerson Oct 22 '12 at 22:04
Note also that if $f$ is bounded, then $\sum fg$ converges. – Gerry Myerson Oct 22 '12 at 22:06
But can we use derivatives if $f$ and $g$ are smooth? – Annick Oct 22 '12 at 22:14
I can see that $\sum fg$ diverges if $f$ is bounded, but what I really mean is that $f$ is the kind of thing whose values smoothly get bigger and bigger, and $g$ is the kind of thing whose values smoothly get smaller and smaller. (I know $g$ could still diverge under that condition, but it doesn't.) And I thought derivatives could be used to establish that $f$'s values get bigger "more powerfully" than those of $g$ get smaller, which would mean that $\sum fg$ diverges. – Annick Oct 22 '12 at 22:30
No, $\sum fg$ converges if $f$ is bounded. Also, if $f$ gets bigger and $g$ gets smaller, then $f'(x)\gt0\gt g'(x)$, so your "if $f'(x)\lt g'(x)$" hypothesis is never met. More generally, I suggest you look at some examples $f(x)=x^a$, $g(x)=x^{-b}$, to see what kinds of behavior can occur, and what conditions you really mean to impose on the derivatives. – Gerry Myerson Oct 22 '12 at 22:47

Will $\sum_{x=1}^nf(x)\cdot g(x)$ converge or diverge?

It depends. Following Gerry Myerson's suggestion, consider the example $f(x)=x^a$ and $g(x)=x^{-b}$ where $a,b>0$. Then $$\sum_{n=1}^\infty f(n)g(n) = \sum_{n=1}^\infty n^{a-b}$$ which converges if and only if $a-b<-1$.

For generally,

  • for any increasing $f$ you can find a decreasing $g$ so that $ \sum_{n=1}^\infty f(n)g(n)$ converges
  • for any increasing $f$ you can find a decreasing $g$ so that $ \sum_{n=1}^\infty f(n)g(n)$ diverges
  • for any decreasing $g$ you can find an increasing $f$ so that $\sum_{n=1}^\infty f(n)g(n)$ diverges

The fourth item would not be true: there are decreasing functions $g$, e.g., $g(x)=1/\sqrt{x}$, such that $\sum_{n=1}^\infty f(n)g(n)$ diverges for every positive increasing function $f$.

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