# 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 $n \geq N$.

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.

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}

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 $$L$$.

• 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?
– SS'
Oct 5 '18 at 2:26
• Just pick any positive $\epsilon<|L|$ and it will work. Just that it might be good to show why it works explicitly though. Oct 5 '18 at 2:30

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.

.

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, $$a_n \neq 0$$