# Prove via contraposition that if $(a_n)$ is decreasing and $\sum_{n=1}^\infty a_n<\infty$, then $\lim_{n\to\infty}na_n=0$

## "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?

• The negation of $\lim_{n\to\infty}n\cdot a_n=0$ ... either $\lim_{n\to\infty}n\cdot a_n$ exists and is nonzero, or $n\cdot a_n$ does not converge. Commented Apr 14 at 13:57
• @GEdgar isn't that $\exists \varepsilon>0 \; \forall N\in\mathbb{N} \; \exists n \ge N\colon \; n\cdot a_n\geq\varepsilon$ ? Commented Apr 14 at 14:02
• @Olivier04 Yes, I believe it is. In this setting, since the $(a_n)$ are decreasing and non-negative, convergence of $(na_n)$ is equivalent to $\limsup_n na_n = 0$, and what you wrote gives the negation of that. Commented Apr 14 at 14:58

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.

• Since $(a_n)$ is decreasing, you have $a_k\geqslant a_{\sigma(n)}$ when $k\leqslant\sigma(n)$. For the second question, suppose you have $\sigma$ as in my answer, then you can define $\sigma'$ such that $\sigma'(0)=0$ and $$\sigma'(n+1)=\min\{ k>n,\sigma(k)\geqslant 2\sigma'(n) \}$$ then, up to replacing $\sigma$ by $\sigma'$, you can suppose that $\sigma(n)\geqslant 2\sigma(n-1)$. Commented Apr 14 at 14:28
• We have $a_k\geqslant\frac{\varepsilon}{\sigma(n)}$ when $k\leqslant\sigma(n)$, this works in particular if $\sigma(n-1)+1\leqslant k\leqslant\sigma(n)$. I added that because I sum over $k\in[\![\sigma(n-1)+1,\sigma(n)]\!]$. Commented Apr 14 at 14:37
• I deleted my comments asking questions, because those are not needed anymore. After you have deleted your answers to those already deleted questions, I will delete this comment. Commented Apr 14 at 15:44
• @Olivier04 I think it's better to let them, it may help someone who reads my answer and has the same questions you asked. Commented Apr 14 at 16:19

This answer is based on https://math.stackexchange.com/a/4899007/1311533

I am just rephrasing in my own words.

It is $$\lim_{n\to\infty}n\cdot a_n\neq0$$, which means $$\exists \varepsilon>0 \; \forall N\in\mathbb{N} \; \exists n \ge N\colon \; n\cdot a_n\geq\varepsilon$$

For every $$N\in\mathbb{N}$$, we are guaranteed to have an $$M\in\mathbb{N}$$ with $$M\geq 2N$$ such that $$M\cdot a_{M}\geq\varepsilon \Longleftrightarrow a_{M}\geq\frac{\varepsilon}{M}$$

Since $$(a_n)_{n\in\mathbb{N}}$$ is monotonically decreasing, we have $$\sum_{k=N+1}^{M} a_k \geq (M-N)\frac{\varepsilon}{M}\geq \frac{1}{2}\varepsilon$$

Thus, $$\sum_{n=1}^\infty a_n$$ is not convergent.