TITLE: Positive Integer Multiple of Convergent, Decreasing Sequence Can Be Made Arbitrarily Small
QUESTION [5 upvotes]: I'm working on the following problem, and for some reason I can't seem to grasp it intuitively:
Let $\{ a_n \}_{n=1}^{\infty}$ be a strictly decreasing sequence of positive numbers. Assume $\sum_{n=1}^{\infty} a_n$ converges. Prove that for every $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that for all $n > N$, $(n-N)a_n < \epsilon$.
Is it correct to say that we are proving that $a_n$ eventually becomes so small that even a large number of copies of the $n^{th}$ term is arbitrarily close to 0?
My attempt:
First, note that as $\sum_{n=1}^{\infty} a_n$ converges, it must hold that the sequence of partial sums converges, and thus is Cauchy. That is, if $s_n = \sum_{k=1}^{n} a_k$, then for fixed $\epsilon > 0$, $\exists N$ such that
$$ |s_n - s_m| < \epsilon \qquad \forall n, m > N.$$
Equivalently, w.l.o.g. assume that $n > m$. Then, since $a_n$ is a sequence of positive numbers, 
\begin{align*}
|s_n - s_m| &= \bigg| \sum_{k=1}^n a_k - \sum_{k=1}^m a_k \bigg| \\
&= \sum_{k=m+1}^n a_k \\
&< \epsilon \qquad \forall n > m > N \\ 
\end{align*}
Now, since $a_n$ is decreasing, 
\begin{align*}
(n-N)a_n &= \sum_{N+1}^n a_n \\
&= a_n + a_n + \dots + a_n \\
&< a_{N+1} + a_{N+2} + \dots + a_n \qquad n>m\geq N+1 \\
&= \sum_{k=N+1}^n a_k \\
&< \epsilon  \qquad \forall n > N \\
\end{align*}
Since $\epsilon > 0$ was arbitrary, $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ such that 
$$ (n-N)a_n < \epsilon \qquad \forall n > N.$$
Is this a valid proof? I feel as if I might be overcomplicating things.

REPLY [2 votes]: Your proof is fine.
One minor nit.
You've used that $s_n$ is Cauchy to say there must be an $N$ such that if $n>m>N$ then $s_n-s_m=\sum_{i=m+1}^{n} a_i<\epsilon.$
That's fine.
But then you assert that $\sum_{i=N+1}^{n} a_i<\epsilon.$ That is not true, because we do not know that $s_n-S_N<\epsilon,$ since it was only true if $m>N.$
You actually need to use $N'=N+1$ rather than $N.$ Then if $n>N'$ you can use $m=N'=N+1$ to get:
$$(n-N')a_{n}\leq s_{n}-s_{m}<\epsilon.$$
Now this is true because $n>m=N+1>N.$

A more direct approach.

If $s_n=\sum_{i=1}^{n}a_i$ and $S=\sum_{i=1}^{\infty} a_i$ then we definitely must have an $N$ so that:
$$\epsilon> S-S_N=\sum_{i=N+1}^{\infty} a_i$$
Now if $n>N$ then:
$$\epsilon>\sum_{i=N+1}^{\infty}a_i> \sum_{i=N+1}^{n} a_i\geq\sum_{i=N+1}^n a_n= (n-N)a_n.$$