# If the series $\sum_1^\infty a_n$ converges, then so does $\sum_1^\infty \frac{{a_n}}{n}$

The problem:

Show that if $$\sum_{1}^\infty a_n$$ converges and $$a_n ≥ 0$$ for all $$n ≥ 1$$, then $$\sum_1^\infty \frac{a_n}{n}$$ also converges. Is the statement true without the hypothesis $$a_n ≥ 0$$ ?

My attempt:

1) $$1\geq\frac{1}{n}$$, because $$a_n\geq0$$ we have $$a_n\geq\frac{a_n}{n}$$ =>if $$\sum_{1}^\infty a_n$$ converge then $$\sum_1^\infty \frac{a_n}{n}$$ also converges.

2) if $$\sum_{1}^\infty a_n$$ converge then partial sums $$s_n$$ is a bounded sequence, and $$\frac{1}{n}$$ is decreasing and $$\frac{1}{n}\rightarrow0$$ from Dirichlet test we have $$\sum_1^\infty \frac{a_n}{n}$$ converges.

But I'm not sure if this is correct I feel something is wrong.

• Didn't you already have it at $a_n \geq \frac{a_n}{n}$ by comparison? Aug 20 '19 at 15:12
• yes but there we know $a_n\geq 0$ at 2) we don'tknow if $a_n\geq 0$ Aug 20 '19 at 15:13
• Ah, I see your question now. Aug 20 '19 at 15:13

It is correct. I would have used Abel's test to justify the convergence of $$\sum_{n=1}^\infty\frac{a_n}n$$ without the assumption that $$(\forall n\in\mathbb N):a_n\geqslant0$$. It allows you to deduce that, say, $$\sum_{n=1}^\infty\frac{na_n}{n+1}$$ also converges.