Theoretical framework around limits

First we will look at the definition of limits and we will try out a few proof to see how it works. In practice we never come back to the actual definition, but we use theorems and properties derived from that. We will see in a next lesson a few of those properties, how to prove them and how to use them.

Convergence Definition

The theory

Let’s look at the case of a finite limit - we will deal with infinity later.

(By the way, the symbol $\epsilon$ is the Greek letter “epsilon”, and it is often used in mathematics to denote a small positive number. Epsilon is kinda the ancester of the letter “e” in the Latin alphabet, but I digress.)

Ouch. That’s quite a mouthful. Let’s try to understand this definition. The overall idea is that we need to give a meaning to the phrase “getting as close as we want to $L$”.

So let’s imagine two mathematicians looking at the proof, let’s call them Alice and Bob.
Bob will be convinced that the sequence converges to $L$ if, each time he gives Alice a number $\epsilon$, Alice can find a number $N$ such that “the sequence is closer than Epsilon” for all $n > N$. In mathematical terms, that would be when we have $|a_n - L| < \epsilon$.

Their dialogue would look like this:

• Bob: “Can you find a rank after which the sequence is closer than 1 to $L$?”
• Alice: “Sure, here it is.”
• Bob: “Ok, well done. Hmmm - what about closer than 0.1?”
• Alice: “Give me a second… Here it is.”
• Bob: “Wow - okay not bad. Can you do it for 0.000000001?”
• Alice: “Sure, here is one”

If Alice can do this for any $\epsilon$, then Bob is convinced that the sequence converges to $L$. This is the idea behind the definition of convergence.

How it goes with an example

Let’s look at the sequence $a_n = \frac{1}{n}$. We would like to prove that the limit of this sequence is $0$. Let’s look at some cases first, like in the discussion between Alice and Bob - and we will play the role of Alice.

• Bob: “Can you find a rank after which the sequence is closer than 0.1 to $0$?” Alice has to find a rank $N$ such that for all $n > N$, we have $|a_n - 0| < 1$.
• Let’s try $N = 1$.
• $|a_1 - 0| = |1 - 0| = 1$ - not good enough.
• Let’s try $N = 2$.
• $|a_2 - 0| = |0.5 - 0| = 0.5$ - not good enough but getting closer. A quick work of algebra will show us that for $N = 10$, we have $|a_{10} - 0| = |0.1 - 0| = 0.1$. Now Alice has to prove that for every $N > 10$, we have $|a_n - 0| < 0.1$. I will leave this aside for now as we will have to do something really similar for the general case.

Generally speaking we can try a few values of $\epsilon$ to get a feel for the way we can “build” the value of $N$. Let’s look at the general example. Bob asks us to find a rank $N_0$ such that for all $n > N_0$, we have $|a_n - 0| < \epsilon$. Please note that we are looking for one of such $N_0$ - we don’t have to find the minimum or anything like that. Just find one which works! We have $a_n = \frac{1}{n}$, so $a_n < \epsilon$ is equivalent to:

• $\frac{1}{n} < \epsilon$. When we invert both sides of the inequality, we have to reverse the sign of the inequality:
• $n > \frac{1}{\epsilon}$. Now we have a good candidate for $N_0$, we have to prove that the sequences stays close to the limit for every rank above $N_0$.

If $N > N_0$, then we have $\frac{1}{N} < \frac{1}{N_0}$.
We know that $\frac{1}{N_0} < \epsilon$, so we have $\frac{1}{N} < \epsilon$.

Here we go, we have proved that for every arbitrary $\epsilon$, we can find a rank $N_0$ such that the sequence is closer than $\epsilon$ for all ranks above $N_0$. So we can say that the limit of the sequence $a_n = \frac{1}{n}$ is $0$.

Infinite limit definition

The theory

The theory is a similar idea than the one above. But instead of proving that the sequence is “closer” to the limit, we need to prove that the sequence becomes “bigger” than any arbitrary number after a certain rank.

Here again we have to prove the existence of a certain number, so we will need to build it according to the value of $K$.

How it goes with an example

Let’s look at the sequence $a_n = n^2$. We would like to prove that the limit of this sequence is $+\infty$.

Let’s look at some cases first, like in the discussion between Alice and Bob. Can we go higher than, say, $100$?
Well, for $N=10$ we have $a_{10} = 100$.
We know as well that if $a > b > 0$, then $a^2 > b^2$. So we can say that for $N > 10$, we have $a_N = N^2 > 100$.

Looking at the general case, let’s look at a given $K$. We have to find a rank $N_0$ such that for all $n > N_0$, we have $a_n > K$. We have $a_n = n^2$, so $n^2 > K$ is equivalent to:
$n > \sqrt{K}$.

So we can take any $N_0$ bigger than $\sqrt{K}$, and we are done. Every $n > N_0$ will have $n^2 > N_0^2$ and we know that $N_0^2 > K$. So we can say that for every $n > N_0$, we have $a_n > K$. We have proved that the limit of the sequence $a_n = n^2$ is $+\infty$.

The negative case

Having a limit of $-\infty$ is the same idea, but we have to prove that the sequence is smaller than any arbitrary number after a certain rank.