If a series converges with decreasing terms, then na_n -> 0

[Mr Davis 97]

Homework Statement

Prove that if $(a_n)$ is a decreasing sequence of positive numbers and $\sum a_n$ converges, then $\lim na_n = 0$

Homework Equations

The Attempt at a Solution

Let $\epsilon >0$. By the Cauchy criterion there exists an $N\in \mathbb{N}$ such that $\forall n\ge m\ge N$, we have that $|\sum_{k=m+1}^{n}a_k|<\epsilon$. But the sequence is decreasing, so $|(n-m)a_n|\le |\sum_{k=m+1}^{n}a_k|<\epsilon$. So we have that $|na_n-ma_n|<\epsilon$ for all $\epsilon>0$. So $na_n=ma_n$ for all $n\ge m\ge N$. Since the tails of these sequence are the same eventually, they have the same limit. Since $\sum a_n$ converges, we see that $\lim ma_n = 0$. Hence $\lim n a_n = 0$.

 

[LCKurtz]

That argument doesn't work. Given $\epsilon > 0$ there exists $N$ such that...
The problem is that the $N$ depends on $\epsilon$. So the (obviously false) conclusion that $na_n = ma_n$ for all $n\ge m$ doesn't follow. Try again.

 

[Mr Davis 97]

That argument doesn't work. Given $\epsilon > 0$ there exists $N$ such that...
The problem is that the $N$ depends on $\epsilon$. So the (obviously false) conclusion that $na_n = ma_n$ for all $n\ge m$ doesn't follow. Try again.

I guess that conclusion is obviously false when I look at it now... So I have that $|(n-m)a_n|<\epsilon$. How can I get rid of that $m$ to get my result? Is there something simple that I am overlooking?

 

[LCKurtz]

Apparently it is possible to modify your argument to make it work. There is a long discussion about this problem at:
https://math.stackexchange.com/questions/4603/series-converges-implies-limn-a-n-0
so I will refer you to it.