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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 $\sum\limits_{n=1}^{\infty} \frac{1}{n} r_{n}$ converges.

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

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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
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
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, 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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