Can the archimedian postulate prove the limit of the series $1/n$ is $1$?

Question

So I know that the sequence $a_n = 1/n$ will converge to $0$ as $n$ approaches infinity. By the definition of a limit this means that for all $\epsilon>0$ there exists $N$ in the naturals such that for all $n\geq N$, $|1/n - 0|<\epsilon$ and this can be proved using the Archimedian postulate, since a $N$ can be found s.t $N>1/\epsilon$ this is all well and good and makes sense.

But the problem comes up if I were to guess the limit wrong and say tried to prove that $1/n\to 1$ as $n$ approached infinity. This would mean showing $|1/n - 1|<\epsilon$ which can also be done using the Archimedian postulate as an $N$ can be found s.t $N>1/(\epsilon+1)$. I don't understand how they can both be proved as the second is clearly untrue.

Answer 1

1. This is the sequence, not the series. (These are two different things).
2. Let's do as you suggest: fix any $\varepsilon > 0$. Indeed, as you point out there exists an integer $N$ such that, for all integers $n\geq N$, $$ n \geq N > \frac{1}{\varepsilon+1}\,. $$ This is true. But what does that give? It is equivalent to $$ \forall n \geq N, \qquad \frac{1}{n} < \varepsilon+1 $$ itself equivalent to $$ \forall n \geq N, \qquad \frac{1}{n} - 1 < \varepsilon $$ This is true, but is not the same as $$ \forall n \geq N, \qquad \left\lvert \frac{1}{n} - 1\right\rvert < \varepsilon $$ since $\frac{1}{n} - 1$ can (and will) be negative...

In short: you forgot to worry about the absolute values.

Answer 2

You appear to have used the supposed identity $|1/n-1|<\epsilon\iff|1/n|<\epsilon+1$

But this is not a correct identity because the absolute value of $1/n-1$ measures that $1/n$ moves away from $1$ as $n$ increases, despite $1/n-1$ itself being negative for all $n>1$.

Therefore pulling the $-1$ out of the absolute value function exchanges the sign of the part within, so for the identity to hold you would need to exchange the signs elsewhere.