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The number $e$ is sometimes defined as $\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^n$. Does this mean we can say that $e = \left(\frac{\infty + 1}{\infty}\right)^{\infty}$? Why or why not? And what is the meaning of $\infty$ here, countable or uncountable infinity?

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closed as off-topic by Andres Caicedo, Danny Cheuk, Andrey Rekalo, rschwieb, Tom Oldfield Aug 6 '13 at 20:10

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I have one thing to add in the hopes that it's illuminating. The issue at stake in this conversation, "what do limits mean?", is one that mathematics as a whole struggled with deeply from the 1680s until around the 1820s-60s. The explanations everybody has been offering in this conversation represent the "current state of the art" which is the definition that was first hinted at by Cauchy in like the 1820s or so and was made precise by Weierstrass 30-40 years later. We tend to talk about this as though it had always been the definition, but the truth is that this definition is the cont'd... –  Ben Blum-Smith Jun 16 '13 at 17:10
    
product of at least a half-century of work attempting to answer the question "how can we define limits without talking about $\infty$ (since in this context we don't have a good definition of $\infty$)?" I learned this history from William Dunham's delightful book The Calculus Gallery, which I highly recommend. It frankly blew my mind. –  Ben Blum-Smith Jun 16 '13 at 17:19

5 Answers 5

Countable infinity and uncountable infinities are sizes of sets.

The "$\infty$" in expressions like "$x\to\infty$" and "$n\to\infty$" is a notational convention. It is not meant to indicate the size of a set. It is not meant to indicate a number of any kind. We use the expression "$x\to\infty$" (respectively, "$n\to\infty$") to say roughly that we want the real variable $x$ (resp., the integer variable $n$) to increase without bound, so we can examine the behavior of some function (resp., sequence).

For example, when we say that $$e=\lim_{n\to\infty}\left(\frac{n+1}n\right)^n,$$ we do not mean "replace $n$ with $\infty,$ and we get $e$." Rather, we mean that there is a unique real number $L$ such that we can make $\left(\frac{n+1}n\right)^n$ get as close as we like to $L,$ just by making $n$ sufficiently large--that is, $\left(\frac{n+1}n\right)^n$ tends toward $L$ as we let $n$ increase without bound--and we call this $L$ by the name "$e$."

Trying to use $\infty$ as a number introduces many potential issues (which is why we shouldn't do that). Stick to the definition, and think in terms of $\varepsilon$ and $N$, instead.

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Ok. I find $\infty$ to be interesting since our current understanding of it is null. Also, it helps explain my belief that you can't map $\Bbb {Z}$ to all $\Bbb {Q}$. –  Daniel Margolis Jun 15 '13 at 20:25
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Incidentally, you can map $\Bbb Z$ (even $\Bbb N$!) to all of $\Bbb Q.$ Note that every rational number can be written uniquely as a fraction $\frac{p}{q}$ where $p,q$ have no common prime factors and $q$ is a positive integer. The map $f\left(\frac p q\right)=2^p3^q$ is a one-to-one map (by Fundamental Theorem of Arithmetic) from the non-negative rationals into $\Bbb N$! (cont'd) –  Cameron Buie Jun 23 '13 at 11:11
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We can use this to construct a bijection $g:\Bbb N\to\Bbb Q$ as follows. Let $g(1)=0,$ and in general, for $n\in\Bbb N$, let $g(2n)$ be the positive rational $r$ with the $n$th least $f(r)$ value--so $g(2)=1,g(4)=2,g(6)=\frac12,g(8)=3,$ and so on--and let $g(2n+1)=-g(2n)$. (ed ajf) –  Cameron Buie Jun 23 '13 at 11:15
    
@Daniel: Mathematicians understand $\infty$ quite well. –  Hurkyl Dec 4 at 1:41
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@Hurkyl: That's fair. Even in the extended number line, I tend to think of it as a piece of convenient convention, since arithmetic falls apart on the extended number line. I recommended to Daniel on another post that he avoid the extended real numbers at the time, since he was already conflating infinite cardinals with numbers with which one could do real arithmetic. It seemed to me that yet another version would only confuse things further. –  Cameron Buie Dec 4 at 3:03

I think you are thinking of:

$e=\lim_{n \to \infty}(1+\frac{1}{n})^n$

Most people do not actually put the infinity symbols directly in though. It just means that as $n$ becomes a higher and higher number, the limit approaches $e$.

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Let $n$ be a natural number and $$\begin{eqnarray*}\mathcal{R}&=&\{1,2,3,\ldots,n\}\\ \mathcal{T}&=&\{0,1,2,3,\ldots,n\}\end{eqnarray*}$$ As we increase $n$, $\mathcal{T}$ will always contain one more element than $\mathcal{R}$. In particular, $\mathcal{R}$ is a subset of $\mathcal{T}$, and the percentage of elements in $\mathcal{T}$ which are also elements of $\mathcal{R}$ is simply $$\frac{\left|\mathcal{R}\right|}{\left|\mathcal{T}\right|}=\frac{n}{n+1}$$ It's not hard to see that if we let $n$ go to infinity, $\mathcal{R}$ becomes $\mathbb{N}$ and $\mathcal{T}$ becomes $\{0\}\cup \mathbb{N}$. These sets have the same cardinality. With this in mind, let's consider a different property of $\mathcal{R}$ and $\mathcal{T}$.

The number of $n$-tuples of a set $A$ is the cardinality of the set $A^n:=\left\{\left(a_1,\ldots,a_n\right)\mid a_i\in A\right\}$.

$\mathcal{R}^n$ can be viewed as the set of list of $n$ numbers with entries from $1$ to $n$, while $\mathcal{T}^n$ can be viewed as the set of lists of $n$ numbers with entries from $0$ to $n$: $$\begin{array}{rccccclccrcccccl}\mathcal{R}^n=\big\{ & \Box&\Box&\Box&\cdots & \Box &\big\}&&\mathcal{T}^n=\big\{ & \Box&\Box&\Box&\cdots & \Box &\big\} \\ &&&&&&&&& \color{red}{0} & \color{red}{0} & \color{red}{0} & & \color{red}{0}\\ & 1 & 1 & 1 & & 1 &&&& 1 & 1 & 1 & & 1\\ & 2 & 2 & 2 & & 2 &&&&2 & 2 & 2 & & 2\\& \vdots & \vdots & \vdots & & \vdots &&&&\vdots & \vdots & \vdots & & \vdots\\ & n & n& n& & n&&&&n & n& n& & n\end{array}$$ Every time $n$ increases by $1$, we gain a new slot in each set of lists, as well as another number to choose from. In the limit, $\mathcal{R}^n$ becomes $\mathbb{N}^\mathbb{N}$ and $\mathcal{T}^n$ becomes $\left(\left\{0\right\}\cup\mathbb{N}\right)^\mathbb{N}$, both of which end up have the same cardinality: they are uncountably infinite, the same cardinality as the real numbers $\mathbb{R}$.

It's easy to see that $\mathcal{R}^n$ is a proper subset of $\mathcal{T}^n$, though, as a tuple is in $\mathcal{T}^n\setminus \mathcal{R}^n$ if and only if it contains a $\color{red}{0}$. It's natural to ask what percentage of $\mathcal{T}^n$ is also contained in $\mathcal{R}^n$. (Note that this is equivalent to the question, "what is the probability that a randomly selected $n$-tuple from $\{0,\ldots,n\}$ does not contain a $0$?")

$$\frac{\left|\mathcal{R}^n\right|}{\left|\mathcal{T}^n\right|}=\frac{n^n}{\left(n+1\right)^n}$$

What happens to this ratio in the limit? Even though we know $\mathcal{R}^n$ and $\mathcal{T}^n$ end up having the same cardinality, their ratio doesn't converge to $1$ as $n$ becomes large. $$\frac{n^n}{\left(n+1\right)^n}\rightarrow \frac{1}{e}$$ In this way, $1/e$ can be viewed intuitively as a ratio of $\mathcal{R}$ and $\mathcal{T}$, but we must be careful. This interpretation is deceptive: if, in the same way, we'd defined $\mathcal{T}_n$ as $$\left\{-\left(m-1\right),-\left(m-2\right),\ldots,0,1,2,3,\ldots, n\right\},$$ we'd have found that $\left|\mathcal{R}^n\right|/\left|\mathcal{T}^n\right|$ would converge to $1/e^m$ (why? consider $\mathcal{T}$ and $\left\{-1\right\}\cup\mathcal{T}$) - yet no matter which $m$ we choose, $\mathcal{R}$ and $\mathcal{T}$ would end up with the same, uncountable cardinality. Thus $1/e^m$ is a only a description of the ratio of the cardinality of these specific sets as $n$ goes to infinity. We must resist the temptation to abuse notation, as writing $$\frac{\left|\mathbb{N}^\mathbb{N}\right|}{\left|\left(\left\{0\right\}\cup\mathbb{N}\right)^\mathbb{N}\right|}=\frac{1}{e},$$ leads us to such falsehoods as $$\left|\left(\left\{0\right\}\cup\mathbb{N}\right)^\mathbb{N}\right|=\left|\mathbb{N}^\mathbb{N}\right|\hspace{20pt}\Rightarrow\hspace{20pt}\frac{1}{e}=1.$$ This is why writing $\infty$ as a part of algebraic expressions is discouraged. Careful thought must be given to what exactly is happening when manipulating infinite sets and their cardinalities.

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Using $\infty$ in a computational expression is bound to cause you confusion. The reality is that the limit of $\left(\frac{n+1}{n}\right)^n$ as $n$ goes to infinity is $e$. But this is never written by any serious mathematician as $\left(\frac{\infty+1}\infty\right)^\infty=e$.

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Don't treat the $\infty$ in $x \to \infty$ as a number. The reason it resembles $x \to c$ is because of a convenient abuse of notation.

The $\epsilon$-$\delta$ definition of a limit is changed slightly when dealing with "limits at infinity". $\lim_{x \to \infty} f(x) = L$ means $\forall \epsilon > 0 \ \exists N \ x > N \implies |f(x) - L| < \epsilon$. This definition contains no infinities, so it can be handled just fine.

In words, for any given tolerance, there is a threshold where anything above it is within said tolerance. Notice the difference from $\epsilon$-$\delta$, where we still need to achieve a certain tolerance, but instead we look at a small neighborhood of $c$.

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