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


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?

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

1 Answer 1


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

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


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

  • $\begingroup$ I see... Thanks. $\endgroup$
    – Analysis
    Aug 4, 2021 at 1:39
  • $\begingroup$ @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. $\endgroup$ Aug 4, 2021 at 1:45

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