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Let’s Get Real… Analysis (Part 2): Sequence Limits

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Sequences are absolutely fundamental to real analysis. They are the bed rock of most of what we will learn that will follow. Not only will our work on infinite series be based on taking limits of sequences, but a lot of definitions in real analysis will have a definition with a similar form. That is, full of \varepsilon's and \delta's.

And wait there’s more!

When you move onto more complex subjects, pun intended, such as complex analysis many of the definitions and theorems are modeled on those in real analysis. Therefore, it is justifiable to spend some time to master these topics.

Enough stalling, let’s get started!

Intuition

What’s a Sequence?

Let’s begin by defining what we mean by a sequence. Loosely we can think of a sequence as an ordered list of real numbers.

Definition (Sequence): We define a sequence, denoted by (a_n) or just a_n, to be a (countably) infinite ordered list of real numbers. (see this footnote for a more rigorous definition)1

Since the list is ordered, we choose to index each term in the sequence by the subscript n\in\mathbb{N}. This allows us to say things like, the first number in the list is denoted by a_1 and the 100th is given by a_{100}, and so on.

Note a sequence is ordered which is different than an infinite set of numbers, since sets are not ordered.

Why’s that a Limit?

Let’s consider the sequence given by \frac{n}{n+1}. The first six numbers in the sequence are:

\frac{1}{2},\;\;\frac{2}{3},\;\;\frac{3}{4},\;\;\frac{4}{5},\;\;\frac{5}{6},\;\;\frac{6}{7},\;\;\cdots

Does it seem like the numbers in the sequence are ‘converging’ to, or ‘approaching’, some number?

It seems like these numbers are approaching 1 in some sense. We can see that they are getting closer and closer and closer to 1 as you go further into the sequence (by further I mean as you look at numbers in the sequence with a larger index. For example, a_{143} is further into the sequence than a_{26} since 143 > 26.).

However, it’s also true that the numbers in the sequence are getting closer and closer and closer to 2 as you go further into the sequence. Since every number in the sequence is increasing, they are all indeed getting closer to 2. But we know that this is not what we meant when we said the numbers in the sequence are approaching 1. What are we missing then? We’re missing that the numbers in the sequence are not just getting closer and closer to 1 as you go further into the sequence, it’s that they are getting arbitrarily close to 1 as you go further into the sequence.

Let me repeat for emphasis,

The items in this sequence get arbitrarily close to the number 1 as you go further into the sequence.

In a moment, we will say the limit of the sequence \left(\frac{n}{n+1}\right) is equal to 1, for this exact reason. In fact, we can think of this statement as a non-rigorous definition for what it means for a sequence to have a limit.

Non-Rigorous Limit Definition: We say that a sequence a_n has a limit \alpha if the items in the sequence get arbitrarily close to the number \alpha as you go further into the sequence.

We want to capture these observations that we just made in our rigorous definition of a limit. We want our definition to capture:

  • As you go further into the sequence every number in the sequence is getting arbitrarily close to the limit of the sequence.

To make this quantitative, let’s first denote our sequence (a_n) and its limit \alpha. Now, to capture that the numbers, a_n, in the sequence are getting close to \alpha we need to look at their difference, which is given by: a_n - \alpha. But because it’s easier to focus on positive numbers, (I always loose a minus sign somewhere) we will focus on the absolute value of their difference |a_n - \alpha|.

Next, we want to communicate that the difference |a_n - \alpha| is getting smaller as you go out into the list. That is, as n increases. Let’s do this by introducing some number to stand in for the difference between the terms in the sequence a_n and its limit \alpha. Out of convention we call this \varepsilon. Moreover, because we wanted to focus on only positive numbers we have to have \varepsilon>0.

If the difference |a_n - \alpha| goes to zero, we can say that at some point in the sequence we should always have

|a_n - \alpha| < \varepsilon,

for any choice of \varepsilon>0. Can you see why this should eventually happen for any \varepsilon>0? If this did not happen, then for some \varepsilon>0 we’d have |a_n - \alpha| \geq \varepsilon. But then a_n does not get arbitrarily close to \alpha. And we want a_n to get arbitrarily close to \alpha.

Finally, at every step we’ve said something like ‘at some point in the sequence’ or ‘as we go further into the sequence’ we should always have |a_n - \alpha| < \varepsilon. We want to make this rigorous too! The trick is to notice that there will be a cut off term a_N in the sequence, and once you go further into the sequence than this cut off term we always have |a_{N+k} - \alpha| < \varepsilon. And, since we are indexing the numbers in the sequence we can focus on the index of the cut off term! Let’s denote it N. Thus, for every n>N we have |a_n - \alpha| < \varepsilon.

Let’s bring this all together!

Capturing Our Intuition

Now for the main event.

Definition (Sequence Convergence): Let (a_n) be a sequence, then we say the limit of (a_n) as n approaches infinity, (or simply the limit of (a_n)) equals \alpha if for all \varepsilon>0 there exists an N\in\mathbb{N} such that |a_n - \alpha| < \varepsilon for all n>N.

Or, for those who love symbols:

We say \lim_{n \rightarrow \infty}{(a_n)} = \alpha if

\forall\varepsilon>0,\,\exists N\in\mathbb{N}\,:\,|a_n - \alpha| < \varepsilon,\; \forall n>N.

*There are some other notations that are used instead of \lim_{n \rightarrow \infty}{(a_n)} = \alpha, such as: a_n \rightarrow \alpha.*

If the limit of (a_n) exists, then we say that (a_n) converges.

I bet if this is the first time, or maybe even the 100th time you’re seeing this it seems overly convoluted. But it’s exactly what we had said earlier.

Since the numbers, a_n, in the sequence get arbitrarily close to the limit \alpha, it must be the case that at some point in the sequence their difference will end up less than \varepsilon for all choices of \varepsilon>0. The phrase, at some point in the sequence we always have…, became, there exists an N\in\mathbb{N} such that . . . for all n>N. It’s our way of saying, at some point in the list we will always have |a_n - \alpha| < \varepsilon.

Let’s see an example.

Example 1

Let’s prove what we claimed when we were building our intuition. That \lim_{n\rightarrow \infty}\frac{n}{n+1} = 1.

Scratch Work: Remember, we must show that \Big|\frac{n}{n+1}-1\Big| < \varepsilon. So, let’s play around with this inequality:

\left|\frac{n}{n+1}-1\right| = \left| \frac{n-(n+1)}{n+1}\right| \\ \textit{ } \qquad\;\;\;\;\;\;=\;\;\;\, \frac{1}{n+1} \\ \textit{ } \qquad\;\;\;\;\;\; <\;\;\;\;\;\,\varepsilon\qquad \mathrm{we}\;\mathrm{want}\;\mathrm{this}.

Let’s solve for n in terms of \varepsilon.

n > \frac{1}{\varepsilon}-1.

So, if we let N = \lfloor\frac{1}{\varepsilon}\rfloor , then we have what we want! Also, for those who haven’t seen this notation before, \lfloor x\rfloor , is called the floor function. Basically, you round x down to the nearest integer. For example,

  • \lfloor 1.1 \rfloor =1,
  • \lfloor 26 \rfloor =26,
  • \lfloor 10.99999 \rfloor =10,
  • \lfloor -1.1 \rfloor =-2,
  • \lfloor 43.26 \rfloor =43.

All this means that the floor function will satisfy x-1< \lfloor x \rfloor \leq x. Which is exactly what we want!

Let’s write our proof out formally.

Proof: Let \varepsilon>0 and N = \lfloor\frac{1}{\varepsilon}\rfloor. Then, for all n>N we have,

\left|\frac{n}{n+1}-1\right| = \left| \frac{n-(n+1)}{n+1}\right| \\ \textit{ } \qquad\;\;\;\;\;\;=\;\;\;\, \frac{1}{n+1} \\ \textit{ } \qquad\;\;\;\;\;\; <\;\;\;\,\frac{1}{N+1} \\ \textit{ } \qquad\;\;\;\;\;\; =\;\;\, \frac{1}{\lfloor \frac{1}{\varepsilon}\rfloor +1} \\ \textit{ } \qquad\;\;\;\;\;\; < \;\;\;\,\frac{1}{\left( \frac{1}{\varepsilon}\right)} \\ \textit{ } \qquad\;\;\;\;\;\; =\;\;\;\;\;\,\varepsilon

Where we used that x-1< \lfloor x \rfloor \leq x implies 1/( \lfloor x \rfloor +1)<1/ x. We have shown, using the definition of a limit that \left(\frac{n}{n+1} \right) \rightarrow 1.

\square

Example 2:

Let’s try this one on for size:

Find the limit of the sequence \lim_{n\rightarrow \infty}\left(\frac{ 26n^2 + 10n}{n^2}\right) = ???

Scratch Work: First, we must find what the limit is, otherwise we are at a loss for showing that the sequence gets close to it!

There are a few strategies to find a limit: (1) use a computer to graph it; (2) graph it by hand; (3) just find the first few terms; and/or (4) use what you know from calculus to make a guess.

We will do this by method (4), since (1)-(3) require too much work for me at the moment!

Take a look at our expression

\frac{26n^2 + 10n}{n^2}

Let’s simplify this expression by getting rid of as many n as we can.

\frac{26n^2+ 10n}{n^2} = 26 + \frac{10}{n}.

Well, as n grows it makes sense that \frac{10}{n} will get smaller and smaller! So maybe the limit is 26, let’s give that a shot!

\left|\frac{26n^2 + 10n}{n^2}-26\right| = \left| 26 + \frac{10}{n} -26\right| \\ \textit{ } \qquad\qquad\qquad= \frac{10}{n} \\ \textit{ } \qquad\qquad\qquad <\varepsilon\qquad \mathrm{we}\;\mathrm{want}\;\mathrm{this}.

Again, solving for n,

n > \frac{10}{\varepsilon} .

Thus, if we have N = \lfloor\frac{10}{\varepsilon}\rfloor +1 > \frac{10}{\varepsilon} we’re done! Lets write this out.

Proof: Let \varepsilon>0 and N = \lfloor\frac{10}{\varepsilon}\rfloor +1 . Then, for all Let n>N we have

\left|\frac{26n^2 + 10n}{n^2}-26\right| = \left| 26 + \frac{10}{n} -26\right| \\ \textit{ } \qquad\qquad\qquad= \frac{10}{n} \\ \textit{ } \qquad\qquad\qquad < \frac{10}{N} \\ \textit{ }\qquad\qquad \qquad = \frac{10}{\lfloor\frac{10}{\varepsilon}\rfloor+1} \\ \textit{ }\qquad\qquad \qquad < \frac{10}{\left(\frac{10}{\varepsilon}\right)} \\ \textit{ }\qquad\qquad \qquad =\varepsilon.

We have shown, using the definition of a limit that \left(\frac{ 26n^2 + 10n}{n^2}\right)\rightarrow 26.

\square

Pro Tip!

Recall in the definition of convergence that we have, for all \varepsilon>0 there exists … etc. Since we want to emphasize that our proofs are valid for any \varepsilon>0 we start each proof with: Let \varepsilon>0.

Example 3: When a limit does not exist!

Up until now, we have considered only sequences that are well behaved enough to have a limit. This is good pedagogically; however, we should note that most sequences don’t have a limit! For one reason or another, they never settle down enough to converge. We say these sequences diverge. There are three main types of divergences.

  1. The sequence shoots off to infinity, in which case we would write a_n \rightarrow \infty.
  2. The sequence shoots of to negative infinity, we likewise would write a_n \rightarrow -\infty.
  3. Lastly, the sequence bounces around so much that the limit does not exist. In which case we would write: \lim{(a_n)} = DNE.

An example of case (1.) is the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, … The numbers just get bigger and bigger! Can you think of an example of (3.)? I’ll give you a moment!

What about, the classic, a_n = (-1)^n. Let’s write out some of the first few terms.

-1, 1, -1, 1, -1, 1, \cdots

do uou see the issue here?

We want to now prove the obvious, that \lim{(-1)^n} = DNE. But to do so, we must translate our definition of convergence to one of divergence. Since divergence is when the limit does not converge, we can negate our previous definition.2 This means that,

Definition (Sequence Divergence): Let (a_n) be a sequence, then we say (a_n) diverges if for all \alpha \in\mathbb{R} there exists a \varepsilon>0 such that for all N\in\mathbb{N} there is an n>N where we have |a_n - \alpha| \geq\varepsilon.

This is saying that, no matter what you think the limit is and how far you go into the sequence (i.e. no matter what you choose \alpha to be and no matter how large N is) there is always a n>N where a_n that is further than \varepsilon>0 from \alpha. (i.e. there will always be some n>N where we have |a_n - \alpha| \geq \varepsilon ) . The sequence never settles down.

Thus, our task is to find some \varepsilon>0 for which there is always a point in the sequence outside of (\alpha - \varepsilon, \alpha + \varepsilon). But, more often than not, it’s easier to do a proof by contradiction. Assume there is a limit and then use the fact that |a_n - \alpha| < \varepsilon must be true for all \varepsilon>0.

Proof: Let \varepsilon=\frac{1}{2}. Now assume for the hope of a contradiction that (-1)^n does have a limit \alpha. Then, there exists some N\in\mathbb{N} such that

\Big| (-1)^n - \alpha \Big| < \varepsilon = \frac{1}{2}.

for all n>N. Let’s get rid of the absolute value,

-\frac{1}{2} < (-1)^n - \alpha < \frac{1}{2}.

Or,

(-1)^n -\frac{1}{2} < \alpha < (-1)^n +\frac{1}{2}.

(why?) Notice that no matter what N we choose there are always even and odd values of n larger than N. Thus, when n is even we have

\frac{1}{2} < \alpha < \frac{3}{2}.

and when n is odd we have

-\frac{3}{2} < \alpha < \frac{1}{2}.

But we cannot have these two situations since we are concerned with all n>N. Thus, (-1)^n must diverge.

\square

Last Example: When the limit is infinity

Let’s finish today coming up with, and then using once, a definition for what it means when the limit of a sequence is infinity.

Well, before we wanted the sequence to settle down to some value. Now we want the sequence to get arbitrarily large! This must mean that no matter how large of a number you want, call it M, there is a point in the sequence (a cut off in the sequence) so that everything is larger than M beyond this point. Thus, our definition becomes:

Definition (Sequence Diverges to Infinity): Let (a_n) be a sequence, then we say the limit of (a_n) divergres towards \infty if for all M>0 there exists an N\in\mathbb{N} such that a_n>M for all n>N.

Note the subtle changes here. Now, we are saying that for any large number M, at some point (after a_N) we have a_n>M.

Let’s prove that (n) diverges to infinity.

Scratch Work: The idea is the same as before, we want to show n>M, so we let N = \lfloor M \rfloor +1>M.

Proof: Let M>0 and N = \lfloor M \rfloor +1. Then, for all n>N we have

n > N = \lfloor M \rfloor +1>M.

Just like that, we’re done!

\square

Concluding Remarks

There’s so much more we could discuss regarding limits! They are endless fun! However, this article was intended to be an introduction, and I believe this is sufficient for now. Next time, we will learn the famous triangle inequality and discuss how limits are unique. I challenge you to try to prove that if a limit exists, then it has to be unique. If you want a hint, check out this footnote!3

I hope you had some fun, but so that way you can have a kick in the discovery yourself, try some of these out:

  1. \lim_{n\rightarrow \infty}\left(\frac{ 10n + 1}{n^2}\right) = ???
  2. \lim_{n\rightarrow \infty}\left(\frac{ 1}{n^2}\right) = ???
  3. \lim_{n\rightarrow \infty}\left(\frac{ (-1)^2}{n}\right) = ???
  4. \lim_{n\rightarrow \infty}\left(n + \frac{ 1}{n^2}\right) = ???

Footnotes:

  1. We could also define a sequence (a_n) to be a function a from the natural numbers \mathbb{N} to \mathbb{R}. i.e. a:\mathbb{N}\rightarrow \mathbb{R}. ↩︎
  2. Recall that when you negate for all it becomes there exists. Likewise, when you negate there exists it becomes for all. If you are unfamiliar with this, I recommend you check this resource out: Truth Tables | Brilliant Math & Science Wiki. ↩︎
  3. Assume that (a_n) converges to both \alpha_1 and \alpha_2. Then try to show that \alpha_1 = \alpha_2. ↩︎

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