# Math Help - proof for calc 5

1. ## proof for calc 5

Suppose that {a_n} is a divergent sequence. Show that it has a subsequence {a_{n_k}} with 1+ n_k < n_{k+1} which also diverges

can anyone give me a concise proof for this

2. You could try an argument by contradiction. Suppose the result is false, so that every such subsequence converges. In particular, the subsequences $a_{2n}$ and $a_{2n+1}$ (of even-numbered and odd-numbered terms terms in the sequence) both converge, say $a_{2n}\to l_1$ and $a_{2n+1}\to l_2$. The subsequence $a_{3n}$ also converges, and it's easy to see that its limit has to be both l_1 and l_2. So that means that $l_1=l_2$, and the whole sequence has to converge to that limit.