Showing that a_n > 0 given its decreasing and lim a_n=0

[Mr Davis 97]

I'm reading a proof, and it says that since $\lim_{n\to\infty}a_n = 0$ and $a_{n+1}\le a_n$ for all $n$, it must be the case that $a_n \ge 0$.

This seems very obvious, since if $a_n < 0$ for some $n$ then it would decreasing from then on, so there wouldn't be infinitely many elements contained in every $\epsilon$-neighborhood around $0$. But is this sufficient for proof? What would a write up of a "proof" of this simple statement look like?

 

[andrewkirk]

Assume the opposite. Then there's some $a_n<0$. Then show that every subsequent member of the series is at least distance $|a_n|$ from 0. So 0 can't be the limit.

 

[math771]

Yes, your intuition is correct. To prove this statement rigorously, we can use the definition of a limit. Since $\lim_{n\to\infty}a_n = 0$, for any $\epsilon > 0$, there exists a positive integer $N$ such that for all $n \ge N$, we have $|a_n - 0| < \epsilon$. In other words, $a_n$ is within $\epsilon$ distance from $0$ for all large enough values of $n$.

Now, let's assume that $a_n < 0$ for some $n$. This means that $a_n$ is negative, and since $a_{n+1}\le a_n$, this implies that $a_{n+1}$ is also negative. Continuing this pattern, we can see that all the terms $a_{n+1}, a_{n+2}, a_{n+3},...$ will also be negative. But this contradicts our assumption that $\lim_{n\to\infty}a_n = 0$, because there would be infinitely many negative terms in the sequence, and thus the sequence would not be within $\epsilon$ distance from $0$ for all large enough values of $n$.

Therefore, our assumption that $a_n < 0$ for some $n$ must be false, and hence it must be the case that $a_n \ge 0$ for all $n$. This completes the proof.