1
$\begingroup$

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 $x\ge 1$.

  • 0
    I'm sorry, that is a typo, I did mean *converges*.2012-10-23

1 Answers 1

1

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 $f$.