Intuition behind definition of limit of sequences at infinity with example

Question

The definition given is for every $c>0$, there exists an N such that $a_n$>c for all $n>N$

Please could someone explain this is really basic terms as im struggling to get my head around it. What is $c$ and how is it decided?

If we take the following sequence $a_n$=$n^2$

is $n=1,2,3,4,5$ and $N=1,4,9,16,25$ or is it the other way round?

I understand to work out this question i set c>0. I would then get $n^2>c, n>\sqrt c$

Id choose $N>\sqrt c$. Then $n^2>N^2>c$

Where do I go from here? Thank you

Answer 1

You need to start with a value for $c$ say 101. Now you need to find an $N$ which satisfies the condition. We need $N$ such that if $n\ge N$ then $a_n>c$. So we choose N = 11, since $11>\sqrt {100}=10$. Now if $n>N$, $a_n>c=n^2\ge 121>100$.

Answer 2

No matter how large a number you pick ($c>0$), you can go far enough into the sequence (choose $N$) so that the values of the sequence ($a_n$) always are larger than the chosen number ($c$) when you go further into the sequence ($n> N$).

Say that the sequence is $(a_n)_{n=1}^\infty = n^2$. You don't actually choose a $c$-value, it has to work for a general $c>0$, however let us first try if we chose a value. Let us pick the value $c=100$, then if we choose $N = 10$ we get that $a_n>c=100$ for every $n>N=10$, since $n^2>10^2 = 100$.

We can prove that the sequence tends to infinity by using a general $c>0$. Pick $N=\lceil \sqrt{c}\rceil$ (where $\lceil\cdot\rceil$ rounds up to nearest integer), then we have that for any $n>N$ that $a_n=n^2>N^2=\lceil \sqrt{c}\rceil^2\geq c.$

The point is that we may pick $N$ as a function of $c$, $N(c)$, that ensures the sequence is large enough. Now that we have shown that there always exists an $N(c)$ that ensures that $a_n>c$ for all $n>N(c)$ we have shown that the sequence tends to infinity.

Answer 3

In order to verify the convergence of a sequence $(a_n)_{n\geq0}$ to a certain finite number $\alpha$ we use tolerances $\epsilon>0$ and make sure that $|a_n-\alpha|<\epsilon$ for all large enough $n$.

There are sequences $(a_n)_{n\geq0}$ for which we intuitively feel that $a_n\to\infty$ when $n\to\infty$. How can we capture this feeling exactly? There is no such thing as an $\epsilon$-neighborhood of $\infty$. But we can say the following: Confronted with a stake at some point $c$ arbitrarily far out to the right one can show that $a_n>c$ for all large enough $n$, i.e., for all $n$ larger than some bound $N$ which will depend on the position of the stake.

Take your example $a_n:=n^2$. If you put the stake at some $c>0$, say at $c:=10^{30}$, then $a_n=n^2>c$ for all $n>\sqrt{c}$.

If you want an explicit formula for an integer $N$ such that $a_n>n^2$ when $n>N$ then you can choose $N:=\lceil\sqrt{c}\rceil$. But note that in arguments of this sort we are not required to find the optimal $N=N_{\min}(c)$ that does the job.