I need help proving this question:
Suppose lim(n->oo) An = L and let Bn = A2n for all n being Natural numbers. Prove that lim(n->oo) Bn = L
More generally, if a sequence converges to L, then every subsequence also converges to L. I personally would be inclined to use a direct proof rather than proof by contradiction. Since $\displaystyle \{A_n\}$ converges to L, given any $\displaystyle \epsilon> 0$, there exist N such that if n> N, then $\displaystyle |A_n- L|< \epsilon$. Since $\displaystyle B_n= A_{2n}$, take n> 2N.
(If you do use proof by contradiction, note that you cannot assume that $\displaystyle \{B_n\}$ converges. That is, $\displaystyle \lim_{n\to \infty} B_n$ may not exist.)