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Suppose not. Then for some $ϵ>0$ and for every $N∈\mathbb N$, there exists $n≥N$ such that $|a_n|>ϵ_n$. This is far from enough to conclude that $∑_n|a_n|=∞$. I think there might be counterexamples to the statement. Other than this I don't see what could be useful here.

Note that the monotonicity assumption has been dropped from this classical problem:

Series converges implies $\lim_{n\to\infty}a_n=0$

Suppose $\sum |a_n|$ converges. Is it true that $\lim_{n\to\infty}n|a_n|=0$?

pluton
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Qqq
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    If $\lim_n n|a_n|=a>0$, then $n|a_n|>a/2$ for all large enough $n$. Therefore $|a_n|>\frac{a}{2n}$. Since $\sum_n\frac{1}{n}$ diverges, then so does $\sum_n|a_n|$ – user647486 Apr 19 '19 at 11:12
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    Are you asking about the case $n a_n \to a \neq 0$ or merely $n a_n \not \to 0$? These are different. In the former case you can just do limit comparison with the harmonic series. But in the latter case lots of things can happen. – Ian Apr 19 '19 at 11:18
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    If you take $a_n=0$ for $n$ not a power of $2$ and $a_n=1/n$ when $n$ is a power of $2$, then $\lim_n n|a_n|$ doesn't exist and $\sum_n|a_n|=\sum_k2^{-k}$, which converges. – user647486 Apr 19 '19 at 11:22
  • Oh, sorry i only know the math problem – Qqq Apr 19 '19 at 11:23

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We know that $$ 1+1/2+1/3+\ldots $$ diverges. Now let's apply this to $a_n$. In this case, assume $a_n$ are real and positive. Since $$ \lim_{n\to \infty} na_n=\lambda>0, $$ for sufficiently large $n$, $na_n\geq \frac{1}{2}\lambda, a_n\geq \frac{\lambda}{2n}$.

Since $\sum1/n$ diverges, $a_n$ diverges.

Ma Joad
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