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

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In summary: 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
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Mr Davis 97
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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...
 
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  • #2
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..
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.
 
  • #3
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.
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?
 
  • #4
Yes, looks good. We still don't know on which side of ##L_0## the ##a_n## are, but they are smaller than ##L##.
 
  • #5
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##.
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##?
 
  • #6
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##?
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.
 

What does it mean for a sequence to have a limit?

For a sequence $\{a_n\}$, having a limit $L$ means that as $n$ approaches infinity, the terms of the sequence get closer and closer to $L$. In other words, for any small positive number $\epsilon$, there exists a positive integer $N$ such that for all $n>N$, the absolute value of $a_n-L$ is less than $\epsilon$.

How can you prove that a sequence has a limit?

To prove that a sequence has a limit, you can use the definition of a limit and show that for any small positive number $\epsilon$, there exists a positive integer $N$ such that for all $n>N$, the absolute value of $a_n-L$ is less than $\epsilon$. This can be done by using algebraic manipulations or by using known limits of other sequences.

What does it mean to say that $\lim a_n

Saying that $\lim a_n

How can you use the statement "If $\lim a_n

This statement can be used as a starting point for a proof by contradiction. Suppose that the sequence does not have a limit of $L$, then for any positive integer $N$, there exists a term $a_N$ that is greater than or equal to $L$. However, if $\lim a_n

Is the statement "If $\lim a_n

Yes, the statement is always true. This is because if a sequence has a limit of $L$, then by definition, the terms of the sequence must get closer and closer to $L$ as $n$ approaches infinity. Therefore, there must exist a term $a_N$ that is less than $L$ for some positive integer $N$.

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