Vtyjkyuk

Let $epsilon > 0$

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

This implies $-epsiloncdot n < 1 - nL < epsiloncdot n$

Therefore $1 - nL > -epsiloncdot n$

Therefore $1 < n(epsilon - L)$

Choose $epsilon = 1/n$

Therefore $nL < 2$

Therefore $L < 2/n$ for all $ngeq N$

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?

Since you are attempting to prove, by the definition, that the sequence does not converge (to $L$, when $Lneq0$), you are supposed to fix a $varepsilon>0$. But $varepsilon$ should be a fixed number neverthelsss.

Take $varepsilon=fraclvert Lrvert2$ and let $Ninmathbb N$. You want to prove that there's a $ngeqslant N$ such that $leftlvertfrac1n-LrightrvertgeqslantfracL2$. Just take $n$ large enough so that $ngeqslant N$ and also that $frac1n<fracL2$. Thenbeginalignleft|frac1n-Lright|&geqslantleft|frac1n-|L|right|\&=|L|-frac1n\&>|L|-fracL2\&=fracL2.endalign

$a_n=1/n>0$, 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 $L=0$.