So we have to show the following:
Whether or not the sequence $\{n\}$ converges?
To disprove a statement we need to give one counterexample, but to prove a statement, we need to rigorously prove it right? In this case, I want to use the epsilon definition to show that this sequence does not converge. Would the following work?
Let the sequence converge to $L$. Then there exists $N$, such that for all $n>N$, $n, N \in \mathbb N$, $s_{n} - L < \epsilon$. But how would we move forth from here? Like I am not sure how I would show the contradiction.
Whether or not the sequence $\left\{ \frac{(-1)^{n}}{2n}\right\}$ converges?
This is the same problem I am facing with this question since, even though I want to use the epsilon-delta definition, I can only come up with actual/graphing the sequence as a function on a graphing calculator to show that even though the function oscillates it actually converges for very large values of $n$.
Can someone tell me where my reasoning is wrong/how I can proceed further from where I am currently stuck? Any hints would be great!
As you know, we say that the sequence $s_n$ converges if $$\exists L:\forall \epsilon > 0,\exists N : \forall n \ge N, |s_n - L|<\epsilon$$
If we want to prove that a sequence does not converge we will have to prove the opposite statement: $$\forall L, \exists \epsilon > 0:\forall N, \exists n \ge N, |s_n - L| \ge \epsilon$$
In this case, if you fix $L$ you know that the sequence $s_n=n$ starts approaching $L$, gets close to it but once $n>L$ the sequence keeps going, getting farther and farther away from $L$.
We can encode this statement as follows: we fix $L$ and choose $\epsilon = 1$, for example. Then for all $N$ we must be able to choose $n \ge N$ such that $|n-L|\ge 1$. Well, for $N \le L+1$ we can choose an $n>L+1$, while for $N>L+1$ we can just choose $n=N$ itself.