- #1

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_{n}) is

[itex]\forall \epsilon > 0, \exists N \in \mathbb{N}:n \in \mathbb{N} \geq N \implies d(p_n,p) < \epsilon[/itex]

So, for the experience, I want to derive the limit of the sequence [itex](1/n)_{n \in \mathbb{N}}[/itex]. I would like detailed and harsh criticism on my process. Here are my thoughts:

I want to find an N for each ε such that any natural number greater or equal to N satisfies [itex]-\epsilon < \frac{1}{n} - p < \epsilon[/itex]. This can be rewritten as [itex]p-\epsilon < \frac{1}{n} < p+\epsilon[/itex]. Rewriting the inequality for n instead of 1/n is tricky, but I think this works:

[itex]p-\epsilon < \frac{1}{n} < p+\epsilon \implies

\begin{cases}

\frac{1}{p+\epsilon} < n < \frac{1}{p-\epsilon} & \text{for p ≠ 0 and sufficiently small }\epsilon \\

-\frac{1}{\epsilon} < \frac{1}{\epsilon} < n & \text{for p = 0 and n > 0} \\

n < -\frac{1}{\epsilon} < \frac{1}{\epsilon} & \text{for p = 0 and n < 0} \\

\end{cases}

[/itex]

This works under the assumption that, given p ≠ 0 and ε < |p|, all terms are either positive or negative if p is positive or negative, respectively. Whenever terms of an inequality are of the same sign (positive or negative), their multiplicative inverse also inverts the inequalities. On the contrary, whenever terms of an inequality are of the opposite sign, their multiplicative inverse maintain the same inequality.

Since the goal is to find a N for each ε such that [itex]n \geq N \implies -\epsilon < \frac{1}{n} - p < \epsilon[/itex], n should not be bounded from above lest there be an n > N that does not meet the requirement. [itex]n[/itex] is not bounded above only in the p=0 and n>0 case, and so [itex](\forall ε > 0, \exists N = \left \lceil \frac{1}{ε} \right \rceil: n \geq N \implies -\epsilon < \frac{1}{n} < \epsilon) \implies p=0[/itex].