My understanding from the definition in my book (Rudin) is this.
A seq. $\{p_n\}$ in a metric space $X$ (I only really know $\mathbb R^k$) is said to be a Cauchy sequence if for any given $\epsilon > 0$, $\exists N\in \mathbb N$ such that $\forall n,m\ge N$, $d(p_n,p_m)<\epsilon$.
(1) I see it as, given any tiny value $\epsilon$, we can find a natural number $N$ large enough so that the distance between $p_n$ and $p_m$ is less than $\epsilon$. Am I right ?
The reason I'm asking this is because I was trying to understand the proof of how
$$\sum a_nb_n$$
can converge, and the book said this
$$\left\lvert \sum_{n=p}^{q}a_nb_n\right\rvert \leq \epsilon$$
satisfies the Cauchy criterion and therefore it converges.
I read other questions and answers about the Cauchy sequence, but it didn't really help me…
Can someone explain me what's going on?
Edit:
Suppose
a) the partial sums of $A_n = \Sigma a_n$ form a bounded sequence
b) $b_0 \geq b_1 \geq \dotsb$
c) $\lim_{b \to \infty} b_n = 0$
Using the partial summation formula, algebraically the equation in the bottom is proved
$$\left\lvert \sum_{n=p}^{q}a_nb_n \right\rvert \leq \epsilon$$
Algebraically I had no problem, but I don't know why this proves convergence. I thought to show that a sequence is Cauchy, we need to find the distance between two terms in a sequence. That's where I'm confused.
Best Answer
The rough intuition is that if we go far enough along the sequence we get to a point where it doesn't vary very much. And if that is the case it must stay within a narrow range of values.
If we can reduce the variation arbitrarily (choose $\epsilon$ as small as we like) by going far enough ($N$ terms), then we can narrow the range as much as we like, so that there is ultimately a single value - the limit.
The value of the criterion is that it proves there is a limit without needing to know what the limit is - just using the internal properties of the sequence.