Real Analysis – Prove via Contraposition: If $(a_n)$ is Decreasing and $\sum_{n=1}^\infty a_n<\infty$, then $\lim_{n\to\infty}na_n=0$

Question

"Normal way"

Let $(a_n)_{n\in\mathbb{N}}$ be a monotonically decreasing sequence of non-negative real numbers.
Furthermore, let $\sum_{n=1}^\infty a_n$ be convergent.

To show is $\lim_{n\to\infty}n\cdot a_n=0$

Cauchy's convergence test gives us $$\forall \varepsilon>0 \quad \exists N\in\mathbb{N} \quad \forall m > n \ge N \colon \quad \sum_{k=n+1}^m a_k < \varepsilon$$
Since $(a_n)_{n\in\mathbb{N}}$ is monotonically decreasing, we get $$(m-n)a_m\leq\sum_{k=n+1}^m a_k < \varepsilon$$
Now, let $m\geq 2N$ and $n:=\left\lfloor\frac{m}{2}\right\rfloor$, which satisfies $m > n \ge N$, so we get
$$\frac{m}{2}a_m\leq(m-n)a_m< \varepsilon$$
So, $\lim_{n\to\infty}\frac{n}{2}\cdot a_n=0$ and thus $$\lim_{n\to\infty}n\cdot a_n=0$$

Contraposition

This wasn't a big deal, however I'm wondering, whether the statement can also be proven directly via its contraposition, so:

Let $(a_n)_{n\in\mathbb{N}}$ be a monotonically decreasing sequence of non-negative real numbers.
Furthermore, let $\lim_{n\to\infty}n\cdot a_n\neq0$

To show is that $\sum_{n=1}^\infty a_n$ is not convergent.

By definition, we have $$\exists \varepsilon>0 \; \forall N\in\mathbb{N} \; \exists n \ge N\colon \; n\cdot a_n\geq\varepsilon$$
However, this only gives us $a_n\geq\frac{\varepsilon}{n}$, and then due to $(a_n)_{n\in\mathbb{N}}$ being monotonically decreasing, we get $\sum_{k=1}^n a_k\geq\varepsilon$, and that doesn't seem to be of much use.

Question

Do you see how the proof via contraposition could be done?

Answer 1

Suppose there exists $\varepsilon>0$ and $\sigma:\mathbb{N}\longrightarrow\mathbb{N}$ strictly increasing such that $\sigma(n)a_{\sigma(n)}\geqslant\varepsilon$ for all $n$. Since $(a_n)$ is decreasing, we have $a_k\geqslant\frac{\varepsilon}{\sigma(n)}$ whenever $\sigma(n-1)+1\leqslant k\leqslant\sigma(n)$ so $$ \sum_{k=\sigma(n-1)+1}^{\sigma(n)}a_k\geqslant\varepsilon\frac{\sigma(n)-\sigma(n-1)}{\sigma(n)}=\varepsilon\left(1-\frac{\sigma(n-1)}{\sigma(n)}\right). $$ We can suppose without loss of generality that $\sigma(n)\geqslant 2\sigma(n-1)$, which means that $$ \sum_{k=\sigma(n-1)+1}^{\sigma(n)}a_k\geqslant\frac{\varepsilon}{2} $$ thus $\sum a_n$ can't be finite.