Can the archimedian postulate prove the limit of the series $1/n$ is $1$? 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 $\epsilon&gt;0$ there exists $N$ in the naturals such that for all $n\geq N$, $|1/n - 0|&lt;\epsilon$ and this can be proved using the Archimedian postulate, since a $N$ can be found s.t $N&gt;1/\epsilon$ 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 $|1/n - 1|&lt;\epsilon$ which can also be done using the Archimedian postulate as an $N$ can be found s.t $N&gt;1/(\epsilon+1)$. I don't understand how they can both be proved as the second is clearly untrue.
This is the sequence, not the series. (These are two different things).
Let's do as you suggest: fix any $\varepsilon &gt; 0$. Indeed, as you point out there exists an integer $N$ such that, for all integers $n\geq N$, $$
n \geq N &gt; \frac{1}{\varepsilon+1}\,.
$$
This is true. But what does that give? It is equivalent to
$$
\forall n \geq N, \qquad \frac{1}{n} &lt; \varepsilon+1
$$
itself equivalent to
$$
\forall n \geq N, \qquad \frac{1}{n} - 1  &lt; \varepsilon
$$
This is true, but is not the same as
$$
\forall n \geq N, \qquad \left\lvert \frac{1}{n} - 1\right\rvert  &lt; \varepsilon
$$
since $\frac{1}{n} - 1$ can (and will) be negative...


In short: you forgot to worry about the absolute values.
You appear to have used the supposed identity $|1/n-1|&lt;\epsilon\iff|1/n|&lt;\epsilon+1$

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 $n&gt;1$.

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.