Convergent Series Can Be Bounded by Any ##\epsilon>0##

[Bashyboy]

Homework Statement

Assume that $a_k > 0$ and $\sum_{k=0}^\infty a_n$ converges. Then for every $\epsilon > 0$, there exists a $n \in Bbb{N}$ such that $\sum_{k=n+1}^\infty a_k < \epsilon$.

Homework Equations

The Attempt at a Solution

Since the series converges, the sequence of partial sums must be cauchy. Hence, given $\epsilon > 0$, there exists an $N \in \Bbb{N}$ such that $|\sum_{k=1}^m a_n - \sum_{k=1}^n | < \epsilon$ for every $n,m \ge N$. Letting $n \ge N$ and $m = n + p$, where $p \in \Bbb{N}$ is arbitrary, we get $|\sum_{k=1}^{p} a_{n+k} | < \epsilon$ or $\sum_{k=1}^{p} a_{n+k} < \epsilon$ since the sequence is positive. Since this holds for every $p$, the partial sums are bounded and therefore the series converges to a number less than $\epsilon$; i.e.,

$$\sum_{k=n+1}^{\infty} a_k < \epsilon$$

How does that sound?

 

[mfb]

"Every element in the sequence is smaller than ϵ" is not sufficient to show that the limit is smaller than ϵ.
You don't need the Cauchy criterion, you can directly compare the partial sum to the limit. Alternatively, think about 2ϵ or ϵ/2 somewhere.

 

[Bashyboy]

You are right. I should have begun with $\frac{\epsilon}{2}$, and then my last line would have read "$\sum_{k=n+1}^{\infty} a_k \le \frac{\epsilon}{2} < \epsilon$."