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I'm reading the textbook "Calculus - Early Transcendentals" by Jon Rogawski for my Calculus III university course.

I'm trying for the life of me to understand the wording of this definition, and I wonder if it can be said in simpler terms to get the basic point across.

A sequence $a_n$ converges to a limit $L$, and we write

$$\lim_{n\to\infty} a_n=L$$

if, for every $\epsilon > 0$, there is a number $M$ such that $|a_n - L| < \epsilon$ for all $n > M$. If no limit exists, we say that ${a_n}$ diverges.

It looks like a very straightforward rule, but I just can't make sense of all the variables in the definition.

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3 Answers

up vote 2 down vote accepted

Basically, what this definition says is that no matter how large or small you pick $\epsilon$, there is some index $M$ such that $a_{M+1}, a_{M+2}, \ldots$ are all within $\epsilon$ of a limit $L$.

So, you have an infinite sequence, and it doesn't matter how small you pick $\epsilon$, you can always find an index $M$ such that every term beyond $a_M$ in the sequence is less than $\epsilon$ units away from some number $L$.

Example: Let's say you pick $\epsilon = 0.001$. Well, then I can pick, say, $M = 50$, and every $a_n$ where $n > 50$ is within $\epsilon$ units of $L$. So you pick $\epsilon = 0.0000001$. Then I might have to pick $M = 43578$. No matter how small an $\epsilon$ you pick, I always win the game.

In practice, we don't often compute actual values of $\epsilon$ and $M$ -- most of the time we can't. But we have other tools to show that it is possible to do so (even if we don't know what the values are).

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I think I understand this now. The professor had worded it similar, but I couldn't see the link in his wording to the formal definition. He had stated "A sequence An has the limit L and we write ... if the terms An get as close to L as desired by taking n sufficiently large." –  agent154 Sep 8 '12 at 1:12
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Let’s call $\{a_n,a_{n+1},a_{n+2},\dots\}$ the $n$-tail of the sequence.

Now suppose that I give you a target around the number $L$: I pick some positive leeway $\epsilon$ want you to hit the interval $(L-\epsilon,L+\epsilon)$. We’ll say that the sequence hits that target if some tail of the sequence lies entirely inside the interval.

For instance, if $a_n=\frac1{2^n}$, the $4$-tail of the sequence hits the target $\left(-\frac1{10},\frac1{10}\right)$ with leeway $frac1{10}$ around $0$: the $4$-tail is $$\left\{\frac1{2^4},\frac1{2^5},\frac1{2^6},\dots\right\}=\left\{\frac1{16},\frac1{32},\frac1{64},\dots\right\}\;,$$ and all of these fractions are between $-\frac1{10}$ and $\frac1{10}$.

It’s not hard to see that no matter how small a leeway $\epsilon$ I choose, some tail of that sequence hits the target $(-\epsilon,\epsilon)$: I just have to find an $n$ large enough so that $\frac1{2^n}<\epsilon$, and then the $n$-tail will hit the target.

Of course, in my example the $4$-tail of the sequence also hits the target $\left(0,\frac18\right)$ with leeway $\frac1{16}$ around $\frac1{16}$. However, there are smaller targets around $\frac1{16}$ that aren’t hit by any tail of the sequence. For instance, no tail hits the target $\left(\frac1{16}-\frac1{32},\frac1{16}+\frac1{32}\right)=\left(\frac1{32},\frac3{32}\right)$: no matter how big $n$ is, $$\frac1{2^{n+6}}\le\frac1{2^6}=\frac1{64}\;,$$ so $\frac1{2^{n+6}}$ is in the $n$-tail but not in the target.

When we say that $\lim\limits_{n\to\infty}a_n=L$, we’re saying that no matter how small you set the leeway $\epsilon$ around $L$, the centre of the target, some tail of the sequence hits that tiny target. Thus, $\lim\limits_{n\to\infty}\frac1{2^n}=0$, and $\lim\limits_{n\to\infty}\frac1{2^n}\ne\frac1{16}$: no matter who tiny a target centred on $0$ you set, there is a tail of the sequence that hits it, but I just showed a target around $\frac1{16}$ that isn’t hit by any tail of the sequence.

One way to sum this up: $\lim\limits_{n\to\infty}a_n=L$ means that no matter how small an open interval you choose around the number $L$, there is some tail of the sequence that lies entirely inside that interval. You may have to ignore a huge number of terms of the sequence before that tail, but there is a tail small enough to fit.

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Try to prove this first:

Given $a,b$, then $a=b$ if and only if for very $\epsilon >0$, $|a-b|<\epsilon$.

You have probably heard that we might refer to the distance from a number $a$ to another number $b$ by $d(a,b)=|a-b|=|b-a|$. The definition then says that a sequence of numbers $a_n$ has a limit $L$ if, the distance $|a_n-L|$ can be made as small as we wish "for every $\epsilon >0$" if we seek further in to the sequence "for all $n>M$".

Look at the image. The points are the corresponding values of $a_k$ (don't worry about scaling). Since $|a_n-\mathscr L|<\epsilon$ is the equivalent to $-\epsilon<a_n-\mathscr L<\epsilon$, the idea we want to capture is that we can box the terms of the sequence inside thethe strip of width $2\epsilon$ centered at $\mathscr L$ if we take $n$ large enough. Not that since $\epsilon'<\epsilon$ we need to go further on the sequence, and take an $M'$ greater that the original $M$ that worked before.

enter image description here

Maybe this helps. Consider the sequence $$a_n=\frac{1}{n}$$

If I give you $\epsilon =0.000001$, what $M$ you should choose so that $$\left|\frac{1}{n}-0\right|<0.000001?$$

Do you agree that, in this case $$\lim_{n\to \infty}\frac{1}{n}=0?$$

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Yes - I do understand that the limit of 1/n = 0 in this case. –  agent154 Sep 8 '12 at 1:11
    
@agent154 Well, what about $$\frac{n}{n+1}$$ –  Pedro Tamaroff Sep 8 '12 at 1:33
    
Same thing - (n/n)/((n/n)+(1/n)) becomes 1/1+0 = 1. I do understand how simple limits work; I was just trying to understand what was meant by that verbose rule. –  agent154 Sep 8 '12 at 14:29
    
@agent154 It is not a verbose "rule", it is a formal definition, and you should learn it, understand it, and learn to apply it. It will serve you a lot. What I wanted you to do in the last one is $$\frac{n}{{n + 1}} = 1 - \frac{1}{{n + 1}}$$ and apply your last result. You're not applying the definition in that algebraic manipulation (unless you have proven some algebraic properties about limits of sequences already). –  Pedro Tamaroff Sep 8 '12 at 14:31
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