# Whether or not the sequence $\left\{ \frac{(-1)^{n}}{2n}\right\}$ converges?

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.).

• But then when we start out with the proof we don't know what the limit is right? So then why can we start with 0?
– user832014
Oct 10 '20 at 22:33
• @user00000000001899 We need to guess what the limit is to use the proof by definition.
– user
Oct 10 '20 at 22:34
• @user00000000001899 Guessing a candidate for the limit, and proving that the candidate is indeed the limit are two completely separate parts of such a proof. Only the latter actually requires any form of strict mathematical rigor. For the former, the only rigorous requirement for a valid argument is that it produces a guess at the end. Oct 10 '20 at 22:39
• I see! I tried to do something similar for $\frac{1}{2^{n}}$ but was not able to, since I would need to use ln for this case, but that hasn't been defined. So how would I do it?
– user832014
Oct 10 '20 at 23:02
• In this case we have $1/2^n<\varepsilon \iff 2^n>1/\varepsilon \iff n>\log_2 (1/\varepsilon)$
– user
Oct 10 '20 at 23:09