# $\sum a_k$ convergent and $(a_k)$ decreasing monotonically $\Rightarrow n\cdot a_n\to 0$

$$\sum a_k$$ convergent and $$(a_k)$$ decreasing monotonically $$\Rightarrow n\cdot a_n\to 0$$

Proof

Let $$\varepsilon >0$$. Given the series $$\sum_{k=1}^{\infty}{a_k}$$ converges, there is (using Cauchy's criterion) an Indexx $$n_0$$ such that for all $$m>n>n_0$$ we have:$$0\leq \sum \limits_{k=n+1}^{m}a_k\leq \frac{\varepsilon}{2}$$ Since $$a_k$$ decreases monotonically, we have $$(m-n)a_m<\frac{\varepsilon}{2}$$. $$(*)$$

Choosing $$n_1=2n_0+1$$, $$m>n_1$$ arbitrary and $$n=\left \lfloor \frac{m}{2}\right \rfloor$$, we have $$n>n_0$$ and thus:$$\frac{m}{2}a_m\leq (m-n)a_m<\frac{\varepsilon}{2}$$ Hence we have $$m\cdot a_m<\varepsilon$$ for all $$m>n_1$$ und since $$\varepsilon$$ was chosen arbitarily, we conclude $$m\cdot a_m\to 0$$.

What I don't understand is how we know $$(*)$$ - is this some kind of theorem?

• It's only because $a_m\leq a_k$ for all $k\in \{n+1,..,m\}$ Aug 4, 2021 at 1:41

## 1 Answer

Since $$a_k$$ decreases monotonically, $$a_k\ge a_m$$ for all $$k\le m$$

$$\sum_{k=n+1}^m a_k\ge\sum_{k=n+1}^m a_m=(m-(n+1)+1)a_m=(m-n)a_m$$

ie,

$$(m-n)a_m\le \sum_{k=n+1}^m a_k\lt\frac{\varepsilon}2$$

• I see... Thanks. Aug 4, 2021 at 1:39
• @Analysis: I should nitpick, however, that the Cauchy criterion is usually stated with the strict $\lt\varepsilon$, not that it matters much though, but since you conclude $(m-n)a_m\color{red}{\lt}\frac{\varepsilon}2$, the formulation of the Cauchy criterion should also have the inequality strict. The meat of the proof, however, is the choice of $n_1$ and $n=\lfloor n/2\rfloor$ to construct $ma_m\lt\varepsilon$ for all $m\gt n_1$. You should verify how the choice works to better understand the proof. Aug 4, 2021 at 1:45