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Is there a (real) sequence such that $\sum{a_n}$ diverges but $\sum{na_n}$converges?

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  • $\begingroup$ Not if $a_n \geq 0$ for all $n$. $\endgroup$
    – abnry
    Commented Dec 11, 2013 at 20:44

2 Answers 2

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No, there is no such sequence. Suppose that $\displaystyle \sum_{n=1}^\infty na_n$ converges. Then, since $\{\frac{1}{n}\}_{n\geq 1}$ is monotone and bounded (it goes to zero), we have $\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac{1}{n}na_n$ converging by Abel's test (or by Dirichlet's test); see Abel's test and Dirichlet's test .

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The idea is (see applications of Abel transformation on Wikipedia):

If $\sum_n u_n$ is convergent and $v_n \to 0$ monotonically, then $\sum_n u_n v_n$ is convergent.

Here, take $u_n=na_n$ and $v_n=\frac 1 n$.

So the answer is no.

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