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Suppose that $lim_{n\rightarrow \infty} a_n = L$ and $L \neq 0$. Prove there is some $N$ such that $a_n \neq 0$ for all $n \geq N$.

We know by the definition of convergence of a sequence, $\forall \epsilon > 0, \exists\ N \in \mathbb{N}$ such that $\forall n \geq N$, $|x_n - L| < \epsilon$.

So take such an $N$ which we know exists since we're given the limit $L$. So for the condition $|a_n - L| < \epsilon$ to hold $\forall \epsilon > 0$, $a_n \neq 0$, as otherwise if $a_n = 0$, since $L \neq 0$, we could find an $\epsilon$ such that $\epsilon < |-L| = L$.

Proof seems rather short and maybe not rigorous enough. Wanted thoughts or if there's a better way to do this.

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You might like to choose your $\epsilon$ explicitly might, for example, take $\epsilon = \frac{|L|}2$.

Then we can find $N$ such that for all $n \ge N$, we have $|a_n -L| < \frac{|L|}2$. Hence $$L-\frac{|L|}{2}<a_n< L+\frac{|L|}2$$

If $L<0$, we have $L+\frac{|L|}2=\frac{L}2<0.$ Hence for all $n\ge N$, $a_n<0$.

If $L>0$, we ahve $L-\frac{|L|}2=\frac{L}2>0.$ Hence for all $n\ge N$, $a_n >0$.

Remark:

Be careful that $|-L|$ need not be equal to $L$.

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  • $\begingroup$ Very clear, thanks! On the question of whether my original proof is valid (minus the $|-L| = L$ assumption), as a not so explicit proof, does it still work? $\endgroup$
    – SS'
    Oct 5 '18 at 2:26
  • $\begingroup$ Just pick any positive $\epsilon<|L|$ and it will work. Just that it might be good to show why it works explicitly though. $\endgroup$ Oct 5 '18 at 2:30
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For fun.

Given :

$\lim_{n \rightarrow \infty}a_n=L \not = 0$.

Statement :

There is a $N \in \mathbb{Z^+}$ s.t. for

$ n \ge N$ $a_n \not =0$.

Assume the statement is not true:

For all $N \in \mathbb{Z^+}$ exists a $n \ge N$ s.t. $a_n=0$.

Let $a_{n_k}$ , $k=1,2,3...,$ be a subsequence with $a_{n_k}=0$.

$\lim_{k \rightarrow \infty} a_{n _k}=0$, a contradiction.

.

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WLOG assume $L > 0$, so in the definition of limit, we may take $\epsilon = L$ and then an $N$ can be found so that for all $n \geq N$,

$$ 0 < a_n < 2 L $$

In particular, $a_n \neq 0$

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