# Constructive proof that $\liminf na_n = 0$ if the series $\sum_{n=1}^{\infty} a_n$ converges

As stated in this question If $\sum a_n$ converges then $\liminf na_n=0$, the proof in the title can be proven if it can be shown that there is a subsequence of $$na_n$$ that converges to 0. I was curious if anyone can provide a witness for this statement. The answer given in the linked question is a contradiction argument.

Edit: The sequence $$a_n$$ is positive. I already know that this is true and understand the non-constructive proof for it, but I was wondering if anyone could provide a proof that involves actually constructing a subsequence of $$na_n$$ that converges to 0.

• By "witness" you mean an example? – evaristegd Jul 21 '19 at 7:12
• Maybe not a specific example, but an algorithm for producing such a sequence would suffice – Aphyd Jul 21 '19 at 7:16
• @DonAntonio You only need positive. For if $\liminf_n a_n=\alpha>0$, then $\sum_n a_n\ge c+\frac\alpha2\sum_n\frac1n=\infty$. – Gae. S. Jul 21 '19 at 7:26
• If the series converges, its terms must decrease to $0$. But you do need absolute convergence to be sure this is true, as the alternating harmonic series demonstrates. – Robert Shore Jul 21 '19 at 7:27
• Perhaps my edit will clarify the answer I am looking for. – Aphyd Jul 21 '19 at 7:36

Note that $$\inf_{n+1 \leq p \leq 2n}\,pa_p \leq 2n\inf_{n+1 \leq p \leq 2n}\,a_p \leq 2\sum_{p=n+1}^{2n}{a_p}.$$
As a consequence, $$\inf_{k \geq n+1}\,ka_k \leq 2\sum_{k > n}{a_k},$$ which proves the conclusion.