# Thread:

1. ## Prove a sequence converges to 1

I am new to the world of proofs and I'm working on an assignment for Analysis. While I understand the concept of convergence I'm finding that writing the proofs is difficult for me and I want to ask if my logic is sound here and what I can improve on.

Problem: For each positive integer $n$, let $p_{n} = 1 - \frac{1}{n}$. Show that the sequence $p_{1}, p_{2}, p_{3}, ...$ converges to 1.

Proof:
To show that $p$ converges to 1, we must show that if $S$ is an open interval containing 1, then $\exists$ a positive integer $N$ such that if $n$ is a positive integer and $n >= N$, then $p_{n} \in S$. Let $S$ be an open interval containing 1 and let $n \in \mathbb{Z}^+$and $N \in \mathbb{Z}^+$. Let $S$ be the interval $(1-\epsilon, 1+\epsilon)$, $\epsilon > 0$. Then $|p_{n} - 1| < \epsilon$ in order for $p_{n} \in S$. Since $p_{n} = 1 - \frac{1}{n}$, $|p_{n} - 1| = |1 - \frac{1}{n} - 1| = |-\frac{1}{n}| = |-1 * \frac{1}{n}| = |\frac{1}{n}|$. And $|\frac{1}{n}| < \epsilon$. Since $n$ is positive, we can say $\frac{1}{n} < \epsilon \Longrightarrow \frac{1}{\epsilon} < n$. Now let $N = \frac{1}{\epsilon}$. Since for all $n > N p_{n} \in S$, $p$ converges to 1.

A few things I'm concerned about are at the end when I let $N = \frac{1}{\epsilon}$; if I say this is it implied that $\frac{1}{\epsilon}$ is a positive integer, or can that not be assumed because we don't know that 1 evenly divides $\epsilon$? Also, I showed for $n > N$ instead of $n >= N$ but because of the way I formed the interval I'm not sure how to fix that. Any tips would be greatly appreciated!
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