The discussion revolves around the implications of the limit of a sequence, specifically whether the condition $\lim a_n < L$ guarantees the existence of an index $N \in \mathbb{N}$ such that $a_N < L$. Participants explore the definitions and properties of convergence in relation to this question, without reaching a definitive conclusion.
Participants express differing views on the implications of the limit condition and the existence of specific indices. No consensus is reached regarding the necessity of $N_0$ or its canonical choice.
The discussion highlights the dependence on the definitions of convergence and the role of $\epsilon$ in determining the behavior of the sequence elements. The implications of the limit condition remain unresolved.
List what you have. You wrote ##\lim a_n < L## so this implies: ##(a_n)_{n \in \mathbb{N}}## converges to a certain limit ##L_0## and ##L_0 < L\,.## You cannot know whether ##a_n## approaches ##L_0## from below or above, but you have ##L-L_0## space to find sequence elements in it.Mr Davis 97 said:I have a question, not based on any homework but just based on my own readings. If ##L \in \mathbb{R}## and ##L>0##, and if ##\lim a_n < L##, does there necessarily exist an ##N \in \mathbb{N}## such that ##a_N < L##? How would I prove this if its true? I tried to use the definition of convergence but got nowhere. I guess this statement could also be false but I couldn't find a counter example..
I think I might have something. Suppose that ##\lim a_n = L_0 < L##. Also, let ##\epsilon = L - L_0 > 0##. Then by the definition of convergence, there exists an ##N \in \mathbb{N}## such that ##n > N## implies that ##|a_n - L_0 | < L - L_0 \implies a_n < L##. Does this show what I wanted to show?fresh_42 said:List what you have. You wrote ##\lim a_n < L## so this implies: ##(a_n)_{n \in \mathbb{N}}## converges to a certain limit ##L_0## and ##L_0 < L\,.## You cannot know whether ##a_n## approaches ##L_0## from below or above, but you have ##L-L_0## space to find sequence elements in it.
Great. One more one thing. I showed that there exists ##N## such that ##n>N## implies ##a_n < L##. In the end, I wanted to show that there existed an ##N_0## such that ##a_{N_0} < L##. Is there a canonical choice for ##N_0##, or would it suffice to say that I could choose ##N_0## to be any integer greater than ##N##?fresh_42 said:Yes, looks good. We still don't know on which side of ##L_0## the ##a_n## are, but they are smaller than ##L##.
I'm not sure what you wanted to choose the ##N_0## for. The point is that in every small neighborhood (with diameter ##\varepsilon##) there are all but finitely many sequence elements. How (finitely) many are outside isn't of interest. What's important is, that ##\varepsilon## can be freely chosen as small as ever we want. Thus the amount of elements outside, i.e. ##a_0 ,\ldots , a_{N_0}## depends on the choice of ##\varepsilon## alone. Plus minus a few hundred of them doesn't play a role. So the only thing which could be made better than it is usually done, is to write ##N_0=N_\varepsilon## or ##N(\varepsilon)## because it depends on it.Mr Davis 97 said:Great. One more one thing. I showed that there exists ##N## such that ##n>N## implies ##a_n < L##. In the end, I wanted to show that there existed an ##N_0## such that ##a_{N_0} < L##. Is there a canonical choice for ##N_0##, or would it suffice to say that I could choose ##N_0## to be any integer greater than ##N##?