Example of "convergent" sequences with a new definition

Question

Let $(x_n)$ be a sequence. We say $(x_n)$ converges to $L$

if $\exists$ $a>0$ for which there is some $n' \in \mathbb{N}$ such that $|x_n - L| < a$ for all $n>n'$

I am trying to find sequences which converge under this definition. for example, I claim $x_n = (-1)^n \to 1 $. Lets check it. Take $a = 3$ and $n' = 1$. Then

$$ |(-1)^n - 1| = \begin{cases} 2 \; \; \; n = 2k+1 \\ 0 \; \; \; n = 2k \end{cases} $$

Thus, both cases, we see that $|x_n-1| < a $ for all $n > 1$. Thus, it works!

Actually, any sequence which converges under the $\mathbf{usual}$ definition also converges under this. Is this correct?

Answer 1

That seems correct, though by that definition any bounded sequence $(a_n)_{n\in \mathbb{N}}$ (say bounded by M) converges to any number $L$. Just take $a=M+1+|L|$, $n'=1$. Then if $n>n'$, $|x_n-L|\leq |x_n|+|L| < M+1+|L|=a$.

Answer 2

Hint Try to prove that a sequence converges under this definition if and only if it is bounded.

Your observation is true since convergent implies bounded.