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Does there exist a convergent series $\sum _{n=1}^\infty a_n$ of positive terms such that $na_n $ does not converge to $0$ ? I only know that if such a series exists then the sequence $\{a_n\}$ cannot be decreasing . Please help . Thanks in advance

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    $\begingroup$ hint : $\displaystyle\sum_{n \text{ is a square}} \frac{1}{n} $ $\endgroup$ – reuns Sep 19 '16 at 8:50
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Consider $$ a_n=\left\{\begin{array}{} \frac1n&\text{if $n=2^k$ for $k\in\mathbb{Z}$}\\ \frac1{n^2}&\text{otherwise} \end{array}\right. $$

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  • $\begingroup$ Very good answer. The sequence has to be positive though, this can be done with a minor edit. $\endgroup$ – i707107 Oct 19 '17 at 23:43
  • $\begingroup$ @i707107: fixed $\endgroup$ – robjohn Oct 20 '17 at 0:24
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    $\begingroup$ @i707107: Many times questions that are about non-negative numbers are asked about positive numbers. Often, the proof is more complicated, but no more instructive, if we are required to use positive numbers rather than non-negative numbers. In my edited answer, the $\frac1{n^2}$ draws attention away from the important part of the answer, but adds nothing of value other than to make all the terms positive. $\endgroup$ – robjohn Oct 20 '17 at 0:42
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Yes, there exist such series. One cute example is $\sum_{n=1}^\infty a_n$ where: $$ a_n = \begin{cases} 0, & \text{if the decimal expansion of $n$ contains a $9$} \\ \frac{1}{n}, & \text{otherwise}. \end{cases} $$

See for example this answer for more details.

(If you insist on positive terms, replace $0$ with something tending to $0$ quickly.)

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