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Let $\epsilon > 0$

$|1/n - L| < \epsilon$ for some $n\geq N$. (definition of convergence)

This implies $-\epsilon\cdot n < 1 - nL < \epsilon\cdot n$

Therefore $1 - nL > -\epsilon\cdot n$

Therefore $1 < n(\epsilon - L)$

Choose $\epsilon = 1/n$

Therefore $nL < 2$

Therefore $L < 2/n$ for all $n\geq N$

By Archimedean property, we know that this only holds when $L = 0$. Hence the sequence converges to $0$, by uniqueness of convergence the sequence does not converge to any $L > 0$.

My issue with this proof is that when I choose $\epsilon = 1/n$, technically $\epsilon$ is changing for every choice of $N$. Is this still a valid proof?

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Since you are attempting to prove, by the definition, that the sequence does not converge (to $L$, when $L\neq0$), you are supposed to fix a $\varepsilon>0$. But $\varepsilon$ should be a fixed number neverthelsss.

Take $\varepsilon=\frac{\lvert L\rvert}2$ and let $N\in\mathbb N$. You want to prove that there's a $n\geqslant N$ such that $\left\lvert\frac1n-L\right\rvert\geqslant\frac{|L|}2$. Just take $n$ large enough so that $n\geqslant N$ and also that $\frac1n<\frac{|L|}2$. Then\begin{align}\left|\frac1n-L\right|&\geqslant\left|\frac1n-|L|\right|\\&=|L|-\frac1n\\&>|L|-\frac{|L|}2\\&=\frac{|L|}2.\end{align}

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$a_n=1/n>0$, decreasing, bounded below by $0$, hence convergent.

Since $a_n >0$, $\lim_{n \rightarrow \infty} a_n=L \ge 0$.

Assume $L>0$.

Let $\epsilon >0$.

There is a $n_0$, positive integer, s.t. for $n \ge n_0$

$|1/n-L| \lt \epsilon$, or

$-\epsilon < 1/n -L < \epsilon$

$-\epsilon +L < 1/n$.

Choose $\epsilon =L/2$ (for example),

then $L/2 <1/n$.

Archimedean principle:

There is a $n_1$, positive integer, s.t.

$n_1>2/L$.

For $n > m=: \max(n_0, n_1)$:

$L/2 > 1/m \ge 1/n$,

contradiction.

Hence $L=0$.

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