# Math Help - Limit of A Sequence

1. ## Limit of A Sequence

I am having trouble grasping the concept of a limit in terms of sequences as I am trying to learn sequences/series. I feel this is an important concept to be grasped but the definition of the limit is so abstract I can understand for any epsilion greater than 0, meaning any number someone gives you > 0 exists some N, whenever
|an - L| < epsilion and n > N
I can kinda see where you get the nth term but where does some random N come from where is it determined in this defintion and the last bit I mentioned too is a bit abstract

2. yes, limit is a VERY important topic.

the uppercase N is some natural number and n is some large number bigger than N that satisfies $|a_n-L| < \epsilon$

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the formal defination goes like this:

a sequence $(a_n)$ has a limit L if, for every $\epsilon > 0$ there exists a Natural number N such that $|a_n - L| < \epsilon$ whenever $n > N$

i.e. (a_n) -> $L => \forall \epsilon > 0, \exists N \in \mathbb{N}:$

$|a_n - L| < \epsilon whenever n > N$

essentially what you are doing is making the sequence $(a_n)$ as close to its limit L by taking n large enough. This lowercase n has to be bigger than some uppercase N which is a natural number.

Hope that helps!

if you have any qn post back.

3. Originally Posted by RockHard
I am having trouble grasping the concept of a limit in terms of sequences as I am trying to learn sequences/series. I feel this is an important concept to be grasped but the definition of the limit is so abstract I can understand for any epsilion greater than 0, meaning any number someone gives you > 0 exists some N, whenever
|an - L| < epsilion and n > N
I can kinda see where you get the nth term but where does some random N come from where is it determined in this defintion and the last bit I mentioned too is a bit abstract

You could read the first tutorial at the head of the Calculus page by Krizalid about evaluating limits according to the definition. I think it may help you.

Tonio

4. How would find just some "random" N I understand your first statement of choosing some n close to the limit

5. The N you choose isn't random, it depends on $\epsilon$ -- think of it like this: the smaller $\epsilon$ is, the larger your N will be, since you will need to go "further" in the sequence to get closer and closer to the limit... That is, unless $a_n$ is constant for all n.

6. Hm, thank sort of make sense. Perhaps a basic problem could help me visualize this

7. Originally Posted by RockHard
Hm, thank sort of make sense. Perhaps a basic problem could help me visualize this
Prove, using the definition of limit, that the sequence $a_n=5-\frac{3}{n+10}$ converges to $5$ as $n\to\infty$.

This is a nice simple problem that will test your basic understanding of the concept.

8. Originally Posted by redsoxfan325
Prove, using the definition of limit, that the sequence $a_n=5-\frac{3}{n+10}$ converges to $5$ as $n\to\infty$.

This is a nice simple problem that will test your basic understanding of the concept.

Well you can use the Theorem that says you can treat the function a(n) as a function of x to help determine the limit so

The limit as $x\to\infty$
$5-\frac{3}{x+10}$

Similarly by properties of limits we can do

The limit as for $x\to\infty$
$5$ - $\frac{3}{n+10}$
separately

The limit of 5 is obviously just 5 because it is constant and then the limit of the second portion is as such

$\frac{3}{n+10}$

$\frac{\frac{3}{n}}{\frac{n}{n}+\frac{10}{n}}$

which then each indiviual limit is taking your whole limit is equal to

$\frac{0}{1}$ which is obviously 0

then you can combined the to which is again...obviously the limit of 5 which is 5

Hope I did this right or I feel incredible dumb, I don't if i used the definition properly by the terms discussed above

9. Originally Posted by RockHard
Well you can use the Theorem that says you can treat the function a(n) as a function of x to help determine the limit so

The limit as $x\to\infty$
$5-\frac{3}{x+10}$

Similarly by properties of limits we can do

The limit as for $x\to\infty$
$5$ - $\frac{3}{n+10}$
separately

The limit of 5 is obviously just 5 because it is constant and then the limit of the second portion is as such

$\frac{3}{n+10}$

$\frac{\frac{3}{n}}{\frac{n}{n}+\frac{10}{n}}$

which then each indiviual limit is taking your whole limit is equal to

$\frac{0}{1}$ which is obviously 0

then you can combined the to which is again...obviously the limit of 5 which is 5

Hope I did this right or I feel incredible dumb, I don't if i used the definition properly by the terms discussed above
Your reasoning is all fine, but to use the definition of limit we need to use the $\epsilon-N$ argument.

We want to show that $\forall~\epsilon>0$, $\exists N$ such that $n>N$ implies

$|L-a_n|=\left|5-\left(5-\frac{3}{n+10}\right)\right|<\epsilon$

$\left|5-\left(5-\frac{3}{n+10}\right)\right|=\frac{3}{n+10}$

To solve $\frac{3}{n+10}<\epsilon$, we get $\frac{n+10}{3}>\frac{1}{\epsilon}\implies n+10>\frac{3}{\epsilon}\implies n>\frac{3}{\epsilon}-10$

Now that the algebra is done, the actual proof is quite simple.

Proof: Given an $\epsilon>0$, let $N=\frac{3}{\epsilon}-10$. So for $n>N$,

$\left|5-\left(5-\frac{3}{n+10}\right)\right|=\frac{3}{n+10}<\frac{ 3}{(\frac{3}{\epsilon}-10)+10}=\frac{3}{3/\epsilon}=\epsilon$

and we're done.

Note: Most professors won't make you show the work to actually find what $N$ should be. My experience has been that you can just assert what $N$ should be, and as long as it works then it's fine.

10. Hmm it is slowly starting to make some sense, does the order of this statement

$|{a_n - L}| <\epsilon$

matter? I still making to sense the last part you did using the proof the statements above are slowly starting to formulate, you just find the limit set to

$|{a_n - L}| <\epsilon$

and solve for n and I think for the last bit you plugged in what you solved for n > some epsilon in the last bit

Thanks for your time bro, everyone gets reps and thanks after this

11. Originally Posted by RockHard
Hmm it is slowly starting to make some sense, does the order of this statement

$|{a_n - L}| <\epsilon$

matter?
What do you mean here? If you mean does it matter whether it is $\left|a_n-L\right|<\varepsilon$ or $\left|L-a_n\right\|<\varepsilon$ the answer is no. This is because $\left|a_n-L\right|=\left|-\left(L-a_n\right)\right|=|-1|\left|L-a_n\right|=\left|L-a_n\right|$. In other words, absolute values dont care about signs. If you meant does it matter whether it is $\left|a_n-L\right|<\varepsilon$ or $\left|a_n-L\right|>\varepsilon$ the answer is...absolutely. One is saying you can make $a_n$ arbitrarily close...the other is saying, well that you can make their difference arbitrarily large.

12. Thanks man!
Also for his last part where he I assume is solving for epsilon what does that statement exactly mean, I understand the N = statement, AHHHHHHHH nevermind I understand what he was proving

13. Originally Posted by RockHard
Thanks man!
Also for his last part where he I assume is solving for epsilon what does that statement exactly mean, I understand the N = statement
If I am interpreting your question correctly, you mean "what does his "find the N value" mean?". The concept of $\lim_{n\to\infty}a_n=L$ is that just like was said above I can make $a_n$ and $L$ as arbitrarily close (since it's arbitrary let's call that distance epsilon) by making $n$ sufficiently large. In other words if you give me ANY $\varepsilon>0$ (it really does not matter what it is) I need to find some $N\in\mathbb{N}$ such that $n\ge N\implies \left|a_n-L\right|<\varepsilon$. So if a teacher asked you to prove that $\lim_{n\to\infty}\frac{n}{n+1}=1$ what it would mean is that if I give you ANY $\varepsilon>0$ (think of it like a variable...but one that is fixed under specific cases) you can tell me the EXACT $N$ such that $n\ge N\implies \left|\frac{n}{n+1}-1\right|<\varepsilon$. Does that help?

14. Yes very much so thanks, going for another example to do

15. Can anyone provide me with another example before I deem this thread solved

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