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Let $\{a_n\}$ be a decreasing sequence of non-negative real numbers such that $\lim \inf (na_n)=0$ , then is it true that $\lim (na_n)=0$ ?

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    $\begingroup$ I would answer no by heart :P $\endgroup$ – Paolo Leonetti Aug 16 '15 at 14:56
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    $\begingroup$ Hint: Create a sequence which is constant (or decreasing very very slowly) for long stretches, i.e., $a_k=\cdots=a_{k+r}$. Then we can make $ka_k$ as small as desired, but if $r$ is sufficiently large, $(k+r)a_{k+r}=ka_k+ra_k$ could be as large as you'd want. $\endgroup$ – Michael Burr Aug 16 '15 at 15:04
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    $\begingroup$ More or less, the same idea of my answer ;) $\endgroup$ – Paolo Leonetti Aug 16 '15 at 15:06
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Consider the two functions $1/n$ and $1/n^2$. Clearly we dont care of the definitions of $a_1,\ldots,a_{n_0-1}$, as long as they are in decreasing order. Set $$a_{n_0}=\frac{1}{n_0^2}.$$ Let $n_1$ be the smallest positive integer such that $\frac{1}{n_1}<\frac{1}{n_0^2}$. Then define $a_{n_0+1},\ldots,a_{n_1-1}$ to be on a straight line between $\frac{1}{n_0^2}$ and $\frac{1}{n_1}$.

Similarly define $n_2$ to be the smallest positive integer such that $\frac{1}{n_2}<\frac{1}{n_1^2}$, and accordingly define on a straght line $a_{n_1+1},\ldots,a_{n_2-1}$ (in particular, they are in strictly decreasing order), and so on.

It follows that $\liminf_n na_n=0$ while $\limsup_n na_n=1$.

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  • $\begingroup$ Ha! Excellent answer $\endgroup$ – preferred_anon Aug 16 '15 at 15:07
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For $n=2^{k^2}$ to $2^{(k+1)^2}-1$, let $a_n=\frac{1}{k2^{k^2}}$. Then our sequence is non-increasing and has the desired property. Indeed $na_n$ has limit $0$ along the subsequence with $n=2^{k^2}$, but $na_n$ can take on arbitrarily large values. One can perturb the example a bit if one wants strictly decreasing.

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