I Proving $\lim a_n<L \implies$ There Exists $N\in \mathbb{N}$ s.t. $a_N< L$

[Mr Davis 97]

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...

 

[fresh_42]

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..

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]

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.

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]

Yes, looks good. We still don't know on which side of $L_0$ the $a_n$ are, but they are smaller than $L$.

 

[Mr Davis 97]

Yes, looks good. We still don't know on which side of $L_0$ the $a_n$ are, but they are smaller than $L$.

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]

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 $N$?

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.