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.).

  • $\begingroup$ 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? $\endgroup$
    – user832014
    Oct 10 '20 at 22:33
  • $\begingroup$ @user00000000001899 We need to guess what the limit is to use the proof by definition. $\endgroup$
    – user
    Oct 10 '20 at 22:34
  • $\begingroup$ @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. $\endgroup$
    – Arthur
    Oct 10 '20 at 22:39
  • $\begingroup$ 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? $\endgroup$
    – user832014
    Oct 10 '20 at 23:02
  • $\begingroup$ In this case we have $1/2^n<\varepsilon \iff 2^n>1/\varepsilon \iff n>\log_2 (1/\varepsilon)$ $\endgroup$
    – user
    Oct 10 '20 at 23:09

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