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Intuitively, I believe that the series would converge, as the fact that $\lim _{n\to \infty} a_n = 0$ would cause an infinite number of terms in the series to equal $0$ and leave only a finite number of finite, nonzero terms, which would cause their sum to also be finite.

However, I am having some difficulty finding a way to prove or disprove this claim rigorously. Thus far, the closest I have come to doing so has been the root test.

$$\lim _{n \to \infty} |na_n|^{\frac{1}{n}}= \lim _{n \to \infty} n^{\frac{1}{n}}\cdot \lim _{n \to \infty}|a_n|^{\frac{1}{n}}= 1 \cdot \lim _{n \to \infty}|a_n|^{\frac{1}{n}}$$

Does $\lim _{n \to \infty}|a_n|^{\frac{1}{n}}= 0$? If so, how would I go about showing this to be the case?

Edit: As the commenters below remarked, I realize now that the fact that $\lim _{n \to \infty} a_n = 0$ is not enough for that series to converge. However, if I were to add in the fact that $\displaystyle \sum _{n=1}^{\infty} |a_n|$ converges, would that change the result?

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    $\begingroup$ This is extremely not true. For example $\frac1n$ $\endgroup$ – user223391 Feb 17 '16 at 6:37
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    $\begingroup$ If $a_n = \frac{1}{n^2}$, $\sum na_n$ also diverges. $\endgroup$ – choco_addicted Feb 17 '16 at 6:37
  • $\begingroup$ What if the series $\displaystyle \sum _{n=1}^{\infty}|a_n|$ converges? Does that change the problem? $\endgroup$ – Jess Feb 17 '16 at 6:39
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    $\begingroup$ @Jess the example $a_n = n^{-2}$ still works. $\endgroup$ – Henricus V. Feb 17 '16 at 6:42
  • $\begingroup$ @Jess even in the case of $\sum_{n=1}^{\infty} |a_n|$, if we take $a_n=\frac 1n$, then the series will diverge. $\endgroup$ – Error 404 Feb 17 '16 at 6:43
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This may be disproven by counterexample.

Let $a_n = \frac{1}{n^2}$. Then we have that the sum $$\sum_{n=1}^{\infty} \mid n a_n \mid ~= \sum_{n=1}^{\infty} \mid n \frac{1}{n^2} \mid ~= \sum_{n=1}^{\infty} \mid \frac{1}{n} \mid~ = \sum_{n=1}^{\infty} \frac{1}{n}$$

Where the last term is the harmonic series, which diverges.

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