Answer to Question #161757 in Calculus for Vishal

Question

Let {an}∞n=1 be a convergent sequence with limit L. Then prove that any subsequence of {an}∞n=1 is also convergent and has limit L

Answer 1

Given that $\lim \limits_{n \to \infty} a_n = L$ . This implies that given $\epsilon >0, \exist N \in \mathbb{N} \ni | a_n - L| < \epsilon$ whenever $n \geq N$

Let $a_{n_k}$ be a subsequence of $a_n$ . Since $n_k \geq n$ , we have that $n_k \geq N$

Consider,

$|a_{n_k} - L| = | a_{n_k} - a_n + a_n - L|\\ \leq |a_{n_k} - a_n| + | a_n - L|$

Since every convergent sequence is a Cauchy sequence we have that $|a_{n_k} - a_n| < \epsilon$

Also, since $a_n$ is convergent we have that $|a_n - L| < \epsilon$

So, we have that $|a_{n_k} - L | < \epsilon + \epsilon = 2\epsilon$

By the $K-\epsilon$ Lemma, we have that $a_{n_k}$ is convergent, and converges to L.