Assuming $\sum n a_n$ converges, does it follow that $\sum a_n$ converge?

Question

Let us assume that the series $$\sum n a_n$$ converges.

Does it follow that the series $$\sum a_n$$ converge ?

In case the series were positive, it's trivial using the comparison test. What can we say in the general case ?

Answer 1

Let us rephrase the question as: if we know that $\sum_{n\geq 1}b_n$ is convergent, do we also know that $\sum_{n\geq 1}\frac{b_n}{n}$ is convergent? The answer is yes and the technique for proving it is summation by parts. Let us set $B_n=b_1+b_2+\ldots+b_n$. We have $$ \sum_{n=1}^{N}\frac{b_n}{n} = \frac{B_N}{N}+\sum_{n=1}^{N-1}\frac{B_n}{n(n+1)} $$ where $B_N\to L$ implies that $\frac{B_N}{N}\to 0$ as $N\to +\infty$. It also implies that the series $\sum_{n\geq 1}\frac{B_n}{n(n+1)}$ is absolutely convergent by asymptotic comparison with $\sum_{n\geq 1}\frac{1}{n(n+1)}=1$.

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Slight generalization with the same proof: if $\sum_{n\geq 1}b_n$ is convergent and $\varepsilon>0$, then $\sum_{n\geq 1}\frac{b_n}{n^{\varepsilon}}$ is convergent, too.

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This also has an interesting sort of converse, known as Kronecker's lemma: if $\sum_{n\geq 1}\frac{b_n}{n}$ is convergent, then $\lim_{N\to +\infty}\frac{1}{N}\sum_{n=1}^{N}b_n =0$.