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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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    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. ;-)2011-03-30

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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!2011-03-30
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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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    @Jason Great reference! Thanks!2012-04-11