Yes, I am aware this can be done using the squeeze theorem, but I am trying to use the epsilon definition for a limit, and currently feel a little confused. Can someone please help? This is what I have so far:
To prove : the sequence $\left\{ \frac{(-1)^{n}}{2n}\right\}$ converges
Answer: The epsilon definition of a limit says that a sequence is said to converge to some number $x \in R$ if, $\forall \epsilon>0, \exists{M} \in N$ such that $\forall n \ge M, |x-a_{n} | < \epsilon$.
So in this case, we would like to show there is an $x \in R$ such that we are able to pick a suitable $M \in N$ for each $\epsilon > 0$ such that $\forall n>M$, $ |x-a_{n} | < \epsilon$.
But then what do I do from here on forth? I cannot just plug in values for $\epsilon$ and $M$ right? I mean to disprove something I can come up with an example that doesn't work, but in this case I would have to show that this works for all n. And I am not sure how to do that without using the squeeze theorem?!
We need to guess that $x=0$ and then we have that $\forall \varepsilon>0$
$$\left|\frac{(-1)^{n}}{2n}-0\right|=\frac1{2n}<\varepsilon \implies n>M\ge\frac1{2\varepsilon}$$
which prove that $a_n \to 0$.
Note that when dealing with limits we need to distinguish between two cases:
proof by definition: we need to guess what the limit is and apply the definition to prove or to disprove the assumption;
limit calculation: we apply theorems derived from the definition which allows to determine the limit (squeeze theorem, ratio-test, etc.).