Question

Suppose {pn} is a real sequence such that lim n-> infinity pn < 0. Prove that pn < 0 for all n >=n0 for some n0 element of the natural numbers

Answer 1

\[\left\{ p_{n} \right\}\] n = 1 to infinity
is a real sequence such that
\[\lim_{n \rightarrow \infty}p_{n} <0\]
Prove that \[p_{n} < 0\] for all \[n \ge n_{0}\] for some \[n \in \mathbb{N} \]

(I don't know how to write everything on the same line, the equation editor always forces me to a new line :( )

Answer 2

Since the limit converges to some finite negative value(\(L\)), there exists some positive \(\large ϵ\) such that \(\large |Pn−L|<ϵ\) for all \(\large n \ge n_0\)

Answer 3

Furthermore, since the limit L is strictly negative, there exists an $$\epsilon _0>0$$such that $$|x-L|<\epsilon _0\Longrightarrow x<0.$$ I.e. if x is in the epsilon_0 neighborhood of L, then x must be strictly negative.

Answer 4

What rational posted seems like it would just be the starting point of a proof. I'm sure I would have to do more than write that.

So if we say x is in the epsilon neighborhood, we say x is in the interval (L - epsilon, L + epsilon), correct? So in order for that to mean x < 0, then you're saying L + epsilon is <= 0? Since epsilon is positive, how do we know L is less than epsilon to make that statement true?

Must epsilon always be greater than 0 for these proofs? Just out of curiosity.

(On a side note, can anyone tell me how to use the equation editor without needing a new line each time I try to use it? )

Answer 5

Beginning with "\+(" and ending with "\+)" puts things inline I believe.

Answer 6

(without the plus)

Answer 7

You are correct in saying if x is in the \(\epsilon\) neighborhood of L, that x is in the interval \((L-\epsilon,L+\epsilon)\). In your problem they give you that \(L<0\). So I'm saying that because of L being strictly less than 0, there exists \(\epsilon_0>0\) such that $$x\in (L-\epsilon_0,L+\epsilon_0)\Longrightarrow x<0$$

Answer 8

and yes, \(\epsilon\) generally is always strictly positive, because it represents a distance away from something.

Answer 9

Oh, right, that makes sense. Kinda forgot the point of epsilon in the first place, lol. Okay, so I'm not to be concerned with what epsilon is, just that I can say such an epsilon exists I'm good. Do I need to define what that epsilon is in terms of n or n0 or something, though, or just say it exists?

Answer 10

Just say it exists. The only time you can really come up with a formula is when you have a formula for the sequence itself in terms of n. But in this case we dont.

Answer 11

Alright, makes sense. Thanks again ^_^

Answer 12

Welcome :)