Question

Suppose a_{n} ->a and that a_{n} >= b for each n. Prove that a >= b

Answer 1

probably proof by contradiction, and the definition of
\[\lim_{n\to \infty}a_n=a\]

Answer 2

you need to work directly from the definition of the limit being \(a\)
do you know precisely what it is?

Answer 3

For some arbitrary epsilon > 0, there exists n_{0} element of the natural numbers such that for all n >= n_{0} abs( a_{n} - a ) < epsilon.

Still havent learned how to use the equation editor in a way where I can mix the math font with my own, so hopefully that's readable. Definition of convergence is something like that, correct?

Answer 4

yeah usually what you wrote as \(n_0\) is just a capital \(N\) but that it is right

Answer 5

so if you assume by contradiction \(a<b\) that means that \(a+\epsilon =b\) for \(\epsilon>0\)
i bet you can get a contradiction out of that, and the fact that given \(\epsilon>0\) there is an \(N\) such that for all \(n>N\) you have \(|a_n-a|<\epsilon\)

Answer 6

We've been using n_{0} in class, so just what I am used to. So are you just picking epsilon to be b - a, or is that an actual conclusion from saying a < b?