# Infinite series - Archimedean principle

I am taking my first real-analysis course this year and have a midterm on friday. I have NO idea what is going on in this class.

So if anyone could clarify what the following means. The book starts off with talking about Archimedes and his principle which states:

The value of an infinite series, if it exists, is the number T such that given any rational numbers L and M such that L < T < M, all of the finite sums from some point on will be strictly contained in the interval between L and M.

The book then goes on about talking about series that do not converge (ie diverges and how hard it is to get a value for that).

Then the book has the following definition:

An infinite series converges if there is a target value T so that for any L < T and any M > T, all of the partial sums from some point on are strictly between L and M.

So I have no idea what this means. What does the interval mean? I am more confused about L and M and the interval then I am about the series. Why are they giving me an interval?

-
The correct spelling in this context is principle, not principal: you are talking about a "fundamental law, doctrine, or assumption; a rule of conduct", a noun; principal is an adjective, meaning "most important or consequential"; or a noun meaning "a person in a leading position". Neither of the last two fit here. –  Arturo Magidin Nov 2 '10 at 21:10

What a spectacularly rubbish way to define convergence. Here's how to think of it in a simple, yet rigorous way.

For example, suppose we have a sequence $S_{n} = \frac{1}{2^n}$. We know that $S_{0} = 1, S_{1} = \frac{1}{2}, S_{2} = \frac{1}{4}$, and so on and so forth, just as you would have defined it in Calculus.

We might now be interested in the sum of $\sum_{n=0}^{\infty} S_{n} = 1 + \frac{1}{2} + \frac{1}{4} + ...$.

You may remember that this series will converge to 2. How should we express this analytically?

Since we cannot actually sum up an infinite number of terms, we'll never actually get to 2 if we sum this up by hand. However, suppose I only wanted to get a value CLOSE to 2. How close? Let's say we want our sum to be within 0.5 of 2. That is, we want the sum to be between $2-0.5$ and $2+0.5$. This is the interval $(1.5,2.5)$. I don't have to sum up all of the terms to get in this interval. In fact, I need to sum up only 3 terms: $1 + \frac{1}{2} + \frac{1}{4} = 1.75$ to be in this interval.

Clearly, $\sum_{n=0}^{3} S_{n}$ is in the interval. We can, if we want to, sum up more than 3 terms, and we will still remain in the interval. The point is that any number of terms more than 3 will still keep us in the interval. We can also say that we need at least the partial sum $S_{2}$ (which is our third term).

Now, suppose we wanted our interval to be smaller, between $2-0.01$ and $2+0.01$, which is the interval (1.99,2.01). It turns out you need 8 terms for our series to get within this interval. After these eight terms (i.e. after the partial sum $S_7$), your sequence stays in the interval.

In our above examples, we wanted to get within 0.5 or 0.01 of the point of convergence. These are our epsilons. That is $(2-\epsilon, 2 + \epsilon)$ is our interval, and we can set $\epsilon$ to be anything we like. We chose to set epsilon to 0.5 or 0.01, but we could have chosen a smaller number.And therein lies the idea of convergence.

You see, no matter how small I chose my epsilon (as long as it is greater than $0$), I can still find a point where my sequence enters the interval $(2-\epsilon, 2 + \epsilon)$, and never leaves it afterwards. When epsilon was 0.5, we needed $S_2$, when epsilon was 0.01, we needed $S_7$. If we can demonstrate that no matter HOW SMALL you make your epsilon, there is a $S_n$ that is in that interval (to stay), you have shown convergence.

How would I actually do this? Well, $\sum_{n=0}^{k} \frac{1}{2^n}$ is a geometric series equal to $2^{-k}(2^{k+1}-1)$. There is a formula for this in your textbook, I presume. If we want $2^{-k}(2^{k+1}-1) > 2 - \epsilon$ we can solve algebraically to get that $k> -\frac{\log{\epsilon}}{\log{2}}$. If we set $\epsilon=0.01$, we'll get that $k>6.64...$. This means that we need at least $S_7$ as I mentioned above. Either way, the mere existence of this function demonstrates that we can always find a sufficiently large k.

Why does this suffice, conceptually? In order for you to argue that the sequence doesn't converge to a particular point, you need to show that the sequence is always some distance away from my point. Yet, no matter how small you make that distance, it is not small enough, for the sequence gets into that interval, and remains there.

-
Amazing. thanks! –  Tyler Hilton Nov 3 '10 at 21:12
hey is it possible for you to give an example of a series that does not converge? ie, no matter how small the epsilon we choose, theres always no "delta" (i presume, the delta is the number of terms?) that makes the series go into that interval and stay there? –  Tyler Hilton Nov 3 '10 at 21:18
I really wouldn't use delta when dealing with convergence. Delta isn't the number of terms, but rather an idea related to limits. Divergence doesn't mean that there's no way to get in the interval. Divergence means that there's no point to get close to. Here are two different scenarios of divergent series: 1) Consider the series $S_{n} = n$. This series gets bigger and bigger and never converges to a point. No matter how many terms you use, the sequences gets bigger still. 2) Consider the sequence $S_{n} = (-1)^{n}$. This sequence bounces back and forth between 0 and 1, never converging. –  Isaac Solomon Nov 4 '10 at 5:46
How would you prove that the series Sn = (-1)^n dosnt converge analytically (using algebra?). –  Tyler Hilton Nov 4 '10 at 19:35
There are various theorems that would show $Sn=(-1)^n$ doesn't converge. Simply put: suppose it converges it X. This means it will get very close to X. Suppose $S_{n}$ is within 0.01 of X. Then $S_{n+1}$ will be 1 away, and certainly won't be within 0.01 of X. –  Isaac Solomon Nov 5 '10 at 3:03

"The interval between $L$ and $M$" means all the numbers that are between $L$ and $M$; we call $L$ and $M$ the "endpoints" of the interval. If you imagine the real numbers as the points on the real number line, with $L$ and $M$ located somewhere, the "interval" would be the segment of the line that is between the points $L$ and $M$.

Usually, an interval may or may not contain (either or both) of its endpoints. In your case, you are looking at the interval that does not contain either $L$ and $M$ (that is the "strictly between $L$ and $M$" condition in the second quote).

For concreteness, say $L=0$ and $M=1$. Then the interval of numbers strictly between $L$ and $M$ would be all real numbers that are greater than $0$ and smaller than $1$.

Given a number $x_0$, if $a\lt x_0$ and $x_0\lt b$, then we think of the interval determined by $a$ and $b$ (all real numbers strictly between $a$ and $b$: this includes $x_0$, but will include other things as well) as being a "neighborhood" of $x_0$: they are all points that are within a certain distance from $x_0$, either below $x_0$ or above $x_0$ (no further than $a$ is if they are below $x_0$, and no further than $b$ is if they are above). The definition you quote is basically saying: if you specify how far from $T$ you are willing to tolerate your total (no further than $L$ is from $T$ if the total is less than $T$, and no further than $M$ is from $T$ if the total is more than $T$), then you can always ensure that from some point forward the partial sums will have a total that you is within your margin of tolerance (a total that you can tolerate). This will hold no matter how tolerant or intolerant you are (you are very tolerant if you allow totals that are far away from $T$; that is, both $L$ and $M$ are "far away" from $T$; you are being very intolerant if your $L$ and your $M$ are really, really, really, close to $T$). You can be as intolerant as you wish, but you are required to allow some leeway.

Intuitively, $T$ is what you think the sum of the series "should" be (if it made sense to add infinitely many numbers). The definition says that the sums you get a long the way will get "as close as you care to specify" to that total $T$, provided you add enough terms of your series, and that once you get there, any further refinments (by adding more terms of your series) will still keep you within your specified 'neighborhood' of $T$. It is very similar to the $\epsilon-\delta$ definition of a limit, if you saw that, or to the "intuitive notion" of limit: as long as you stay close enough to the point (for series, as long as you add up enough terms), the value of the function (for series, the running total) will be as close as you care to "pre-specify" from the limit (for series, as close as you care to "pre-specify" from $T$).

Added: To respond to your query in the comments: "If we have a target value, why add the tolerance?" Keep in mind that we can only add finitely many numbers at a time. We can add two numbers, and then using that we can add three, four, five, etc. But we don't really know what it means to add infinitely many numbers, like the series says we "should" or "want". So we cannot actually perform the act of adding up all the (infinitely many) terms. Rather, we can only add up to a certain number of the terms; these are the partial sums. We can (in principle, at least) compute a partial sum to any term we want, but we cannot actually add up all of them (just like we can, in principle, count as high as we want, but we cannot actually count through "all" the natural numbers). So what we have actual access to are the partial sums and only the partial sums. We think that if we could add up all the terms of the series, then it would make sense to call that sum $T$; so what we do is we compare those partial sums to our target, $T$. But of course, $T$ is supposed to be what we would get after adding up all the terms of the series (if we could do that), and what we are comparing is not the result of adding all the terms, but only some of them. So the total we have in hand (the partial sum) has no reason to be exactly the same as the target (just like, if you add the first 100 numbers in a list of a thousand numbers, the running total has no reason to be the same as the final total). So we need to add some tolerance to account for the fact that we are not actually adding up all the terms, but only some of them.

(P.S. I've never seen what you quote as the "Archimedean Principle" called that, but that may very well be just me...)

-
Thanks, I am probably going to have to read that a couple of times. My question is whats the point though? If we have a target value, why add a tolerance in there? –  Tyler Hilton Nov 2 '10 at 21:21
@Tyler: It's a precise way of stating what "approaching the target value" really means. –  Hans Lundmark Nov 2 '10 at 21:46
@Tyler Hilton: I've added an attempt to answer your query in the body. –  Arturo Magidin Nov 3 '10 at 3:38

This is really quite a convoluted way to define the limit of a sequence. I can sympathize with your confusion.

The first thing to note is that the limit of an infinite series is the limit of its sequence of partial sums. So, we just need to understand limits of sequences. The reason for the interval, is that the series may have positive and negative terms. So, while a series is busy converging, it may jump to the left or right of its limit. But the main point is that for any error tolerance (no matter how small) all the partial sums will eventually be within this error tolerance of the limit. Being within a small error tolerance means that you are in a small interval containing the limit (hence the interval business). Here is how we quantify error tolerance and eventually:

A sequence $(s_n)$ converges to a limit $s$ if for any $\epsilon>0$ there exists an index $N$ such that $s_n \in (s-\epsilon, s+\epsilon)$ for all $n \geq N$.

So think of $\epsilon$ as the error tolerance, and the $N$ (which depends on $\epsilon$) as a measure of the eventually. This definition is mathematically equivalent to the one you gave where $s-\epsilon$ corresponds to $L$ and $s+\epsilon$ corresponds to $M$, but this one is superior for a couple of reasons.

First, there is no reason to insist that $L$ and $M$ are rational. We can insist that they are rational because rationals approximate real numbers arbitrarily well, but it is confusing to include it in the definition.

Secondly, there is no reason to specify two error tolerances $\epsilon_1$ and $\epsilon_2$ which is what the confusing definition is doing. Instead of saying we eventually stray by at most $\epsilon_1$ to the left of the limit and by at most $\epsilon_2$ to the right of the limit, we may as well specify a single error tolerance $\epsilon:=\min(\epsilon_1, \epsilon_2)$. We can then insist that the sequence eventually strays by at most $\epsilon$ (in either direction) from the limit.

-
hey, thanks for your time but this still dosn't make any sense. I am not familiar (the prof talks about it) with the epsilon/delta game. –  Tyler Hilton Nov 2 '10 at 21:20
If you are not familiar with "the epsilon/delta game" then you will not understand anything well in the course. It sounds harsh, but that stuff has genuine meaning. You can best understand concepts in analysis by drawing pictures instead of mainly memorizing lines of logic that apparently don't mean anything to you. Learn to understand visually what's happening with limits and then the associated logic can be understood through that. –  KCd Nov 3 '10 at 9:40