Can someone explain a proof that a certain sequence converges by using the Archimedean property, such as the following.
Prove that ${a_n} = \frac{1}{n+1}$ is a sequence that converges to $a=0$ for $n \to \infty$.
Definition: If $a_n$ converges to $a$, then for any $\varepsilon > 0$ there is an index $N$ such that $|a_n - a| < \varepsilon$ for all $n \geq N$.
Proof by the Archimedean property:
Let $\varepsilon >0$ be fixed. By the Archimedean property, there exists an $N$ such that $\frac{1}{N} < \varepsilon$. Since for all $n \geq N$, $a_n = \frac{1}{n+1} < \frac{1}{n} \leq \frac{1}{N}$, we have $|a_n - 0| < \varepsilon$ for all $n \geq N$. Ergo, $\big\{\frac{1}{n+1}\big\}_{n \in \mathbb{N}}$ converges to $0$.
A few questions:
Why is it necessary to state that $n \geq N$?
Since $n$ and $N$ are both suppose to be natural numbers, why is it that Archimedean can’t directly imply $\frac{1}{n} \geq \varepsilon$?
My main confusion is with indexes, if someone could explain to me why we compare an index $n$ to some natural number $N$, that would be nice.
Also, apologies for the poor formatting, I’m still new to stack exchange.
Thank you
I have clarified the proof you gave. Hope that helps you understand the proof.
It seems that you don't fully understand the definition of convergence (or limit of a sequence). The definition of $a_n \to a$ is that, for any $\varepsilon > 0$, we can find a number $N$ such that $|a_n - a| < \epsilon$ for all $n \geq N$. In other words, the sequence converges if the "tail" of $a_n$ gets arbitrarily closer to $a$ as $n \to \infty$.
In order to prove this, we first fix an $\varepsilon > 0$, and then try to find an $N$ so that $|a_n - a| < \epsilon$ for all $n \geq N$. If we can find this number $N$, then the claim in the definition is proven because the $\varepsilon$ we fix is arbitrary. Now the Archimedean property helps us find this $N$.