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Why does $\left(\frac{\infty + 1}{\infty}\right)^{\infty} = e$? Does this account for the disparity between the countable and uncountable $\infty$? Why?

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I voted it down after I read it. –  MJD Jun 13 '13 at 1:30
I think you need to understand that limits are a way to make infinity precise, and that is exactly what you are not doing in your question. –  Javier Badia Jun 13 '13 at 1:40
What does it have to do with your question? Nothing is uncountable there. The infinity in $\infty$ is neither countable nor uncountable. It's not an actual infinity, merely a potential one. –  tomasz Jun 13 '13 at 1:54
Reiterating for emphasis: nobody substitutes the dummy variable of a limit with $\infty$, and in general limits have nothing to do with cardinality (in particular, countable versus uncountable). –  anon Jun 13 '13 at 1:57
Your confusion stems from not understanding what $\lim_{n \to \infty} a_n = L$ means. Look up the precise definition of convergence of a sequence. Observe that the word "infinity" never arises. –  Vectk Jun 13 '13 at 2:05
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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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5 Answers

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
Out of curiosity, what text(s) are you working from? Perhaps if I knew that, it would help me clarify things for you. –  Cameron Buie Jun 16 '13 at 20:27
@Daniel: Have you given up on this question entirely? –  Cameron Buie Jun 23 '13 at 11:06
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
We can use this to explicitly define 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 $g(2n+1)=-g(2n)$. –  Cameron Buie Jun 23 '13 at 11:15
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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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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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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 $0$?")


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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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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