Reason for $|a_n-L| > \frac{1}{i}$, when a sequence is divergent I have a basic question regarding the limit of a sequence.As per the link given below on the section of convergence of sub sequences, when a sequence $\left\{a_n\right\}$ diverges, $\forall i>0$, there is a sufficiently large $a_n$ such that $$|a_n-L|>\frac{1}{i}$$.
https://brilliant.org/wiki/subsequences/
I am actually confused with this because when i take $a_n=(-1)^n$ and $L=-1$ and $i=0.01$, i can choose sufficiently large $a_n$ as $1$ and we get $$|1-(-1)|>\frac{1}{0.01}$$ which is false. So am i missing something?
 A: The statement:

Suppose ... that $\{a_n\}$ does not converge (to $x$). Then, for every $i>0$, there is an arbitrarily large $a_n$ such that $|a_n-x|>\frac1i$.​

is true, with minor clarifications.  But as we'll see, it's not so critical to the proof of the desired claim.
Claim. Suppose that $\{a_n\}$ is a sequence, and $x$ is a real number such that $\{a_n\}$ does not converge to $x$.  Then there exists a positive integer $i_0$ such that for all positive integers $n$ and integers $i > i_0$, there exists an integer $n_i$ such that $n_i > n$ and $$|a_{n_i} - x | > \frac{1}{i}$$
Before the proof, remember the definition of convergence and the proper way to negate quantified statements.  We say that $\{a_n\}$ converges to $x$ if for every $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that for all $n \in \mathbb{N}$, if $n > N$, then $|a_n - x | < \epsilon$.
The negation of this is: $\{a_n\}$ does not converge to $x$ if there exists $\epsilon > 0$ such that for all $N \in \mathbb{N}$ there exists $n \in \mathbb{N}$ such that $n > N$ and $|a_n - x| \geq \epsilon$.
Proof of claim. Since $\{a_n\}$ does not converge to $x$, there exists $\epsilon > 0$ as in the previous paragraph.  Choose $i_0$ to be a positive integer such that $\frac{1}{i_0} < \epsilon$.  Suppose that $i$ is a positive integer greater than $i_0$, and $n$ is any positive integer.  Since $a_n \not\to x$, there exists an integer $n_i > n$ such that $$|a_n - x | \geq \epsilon > \frac{1}{i_0} > \frac{1}{i}$$ QED.
But this is too clever by half.  In order to reach a contradiction, it's sufficient to find an $\epsilon > 0$ and an increasing sequence $\{n_k\}$ such that $|a_{n_k} - x| \geq \epsilon$ for all $k$. This is guaranteed if $\{a_n\}$ does not converge to $x$. Then $\{a_{n_k}\}$ cannot converge to $x$ either.
But even this is unnecessary.  Suppose every subsequence of $\{a_n\}$ converges to $x$.  The sequence $\{a_n\}$ is a subsequence of $\{a_n\}$, so it converges to $x$ as well.
See also A sequence converges if and only if every subsequence converges? from this site.
