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Prove or provide a counterexample:

Let ${r_{n}}$ be a sequence such that $r_{n} \rightarrow 0$ as $n \rightarrow \infty$. Then, the sum $\displaystyle\sum\limits_{n=1}^{\infty} \frac{1}{n} r_{n}$ converges.

This has me stumped... I can't think of any counterexamples, but it just doesn't seem true.

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3  
e.g. $r_n=1/\log(n)$ is a counterexample - compare with $\int 1/(x\log x) dx$. –  user8268 Mar 30 '11 at 21:09
    
Hint: Think about the usual proof that the harmonic series diverges. Then come up with a multiplicative term that goes to zero slowly enough that the proof still works. –  mjqxxxx Mar 30 '11 at 21:10
2  
On the other hand, if $r_n:=\frac{1}{\log^2 n}$ then your series converges (Cauchy condensation test or integral test); but if $r_n:= \frac{1}{\log n\ \log (\log n)}$ your series diverges again; and if $r_n:=\frac{1}{\log n\ \log^2 (\log n)}$ you find convergence again... Therefore, in general, the convergence properties of a series like $\sum \frac{r_n}{n}$ are in no way connected with the rate of convergence of $r_n$ when $r_n$ goes to zero really really slowly. ;-) –  Pacciu Mar 30 '11 at 21:47

2 Answers 2

up vote 6 down vote accepted

This is false. For example, let $r_n = \frac{1}{\ln n}$. Then clearly $r_n \to 0$, and the sum $\sum \frac{1}{n}r_n = \sum \frac{1}{n \ln n}$ diverges.

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Makes sense. Thanks! –  PFHayes Mar 30 '11 at 21:31

Here is a result which shows that there is no smallest divergent nonnegative sequence, nor any largest convergent nonnegative sequence:

  1. Let $(a_n)$ denote a nonnegative sequence such that $\displaystyle\sum_na_n$ diverges. Then there exists a nonnegative sequence $(u_n)$ such that $u_n\to0$ and $\displaystyle\sum_nu_na_n$ diverges.
  2. Let $(b_n)$ denote a nonnegative sequence such that $\displaystyle\sum_nb_n$ converges. Then there exists a nonnegative sequence $(v_n)$ such that $v_n\to+\infty$ and $\displaystyle\sum_nv_nb_n$ converges.
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Do you know of a reference for this fact? –  Jason DeVito Mar 30 '11 at 21:42
    
On second thought, quick googling came up with math.depaul.edu/mash/worstconv.pdf, which is not quite the same (or at least, it's not obvious to me that they're the same), but it's certainly in the same spirit. –  Jason DeVito Mar 30 '11 at 21:54
    
@Jason This is exactly the same result. Good find. So now you do have a proof. –  Did Mar 30 '11 at 22:10
    
@Jason Great reference! Thanks! –  Pedro Tamaroff Apr 11 '12 at 23:09

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