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.

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


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