Real Analysis: Show the sequence $a_n=10n$ is not convergent I am trying to prove this by contradiction so $|10n- λ| \geqslant \epsilon$ but I don’t know where to go from here as in the examples an epsilon is chosen but with no explanation as to why.
 A: In the examples you refer to, an $\epsilon$ is chosen because 


*

*Basically any $\epsilon$ would work

*You only need one to work

*You do need one to work


Given a number $\lambda$, the definition of "$a_n$ converges to $\lambda$" is

For any $\epsilon>0$, there is an $N\in \Bbb N$ such for any $n>N$ we have $$|a_n-\lambda|<\epsilon$$

Negating this, the definition of "$a_n$ doesn't converge to $\lambda$" becomes

There is an $\epsilon>0$ such that for any $N\in \Bbb N$ there is an $n>N$ such that $$|a_n-\lambda|\geq \epsilon$$

So as you can see, as long as they can prove that a single such $\epsilon$ exists, they will have proven that $\lambda$ is not the limit of the sequence. One common way to prove that an $\epsilon$ exists is to pick a specific number and show that that works.
You need to pick an $\epsilon$ which is small enough. If $1$ works, most people would probably pick that. Failing that, usually one would go for $\epsilon = \frac1k$ for some natural number $k$. And if one $\epsilon$ works, then any smaller $\epsilon$ also automatically works.
Some times, I like to pick a smaller-than-strictly-necessary $\epsilon$ in order to ensure that we get $|a_n-\lambda|>\epsilon$ rather than $|a_n-\lambda|\geq \epsilon$, because I like that better (for instance, for $b_n = (-1)^n, \lambda = 0$, then $\epsilon = \frac12$ is good enough but I prefer to use $\frac13$). But you don't need to do that.
