Prove that if <a> converges, then every subsequence of <a> converges and has the same limit as a.
let $\displaystyle \epsilon > 0$
suppose that $\displaystyle \{a_n\} \to a$
let $\displaystyle \{a_{n_k}\}$ be any subsequnce
Since the sequence convereges there is an $\displaystyle N \in \mathbb{N}$ such that for all n > N
$\displaystyle |a_n-a|< \epsilon$
but since $\displaystyle \{a_{n_k}\}$ if futher in the sequnce then
$\displaystyle
\{a_{n}\}
$
so $\displaystyle |\{a_{n_k}\}-a|< \epsilon$
The integer $\displaystyle n_k \ge n$ a subsequence has all of the same terms as the regular sequence. It just skips some
ie $\displaystyle a_n=\frac{1}{n}$ and
$\displaystyle a_{n_k}=\frac{1}{2^{n_k}}$
the first gives the sequence
$\displaystyle a_n=\left\{ \frac{1}{1} ,\frac{1}{2} ,\frac{1}{3} ,\frac{1}{4} ,\frac{1}{5} ,... \right\}$
but
$\displaystyle a_{n_k}=\left\{ \frac{1}{1} , \frac{1}{2} ,\frac{1}{4} ,\frac{1}{8} ,\frac{1}{16} ,... \right\}$
the subsequce skips terms from the original so to get the term
$\displaystyle \frac{1}{16}$ $\displaystyle n=16$ but $\displaystyle n_k=4$
So $\displaystyle n \ge n_k$