Q:

Proving a statement about a sequence (using limit)

Suppose $\{a_n\}$ is a sequence such that $\lim_{n\to\infty} (a_{n+1}-a_n)=0$. 

Prove that if there exists $c>0$ such that for all $n$, $|a_n|\geq c$ then either for all but a finite number of terms $a_n>0$ or for all but a finite number of terms $a_n<0$.

I fail to see how $\lim_{n\to\infty} (a_{n+1}-a_n)=0$ helps here. I know that it means that for all $\varepsilon>0$, eventually $|a_{n+1}-a_n|<\varepsilon$. I tried playing with $\varepsilon$ by setting it to $c$ and use the triangle inequality but with no luck. Any tips?

A:

HINT: If there are infinitely many "changes of signs" (i.e., if the conclusion doesn't hold), you can construct a subsequence of $\{a_{n+1}-a_n\}_n$ which doesn't converges to zero. Use the condition $|a_n|\geq c$.
The best way you can construct this sequence is this: Suppose $a_1$ is negative, let $a_{n_1+1}$ be the FIRST positive number in the sequence after $a_{1}$ (it exists by assumption), then let $a_{n_2+1}$ be the FIRST positive number in the sequence after $a_{n_1+1}$ and so on... argue that $a_{n_k}$ is negative for all $k$, $\{a_{n_k}\}$ will be your sequence. (A similar argument works if $a_1$ is positive).

