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The number $e$ is characterized in calculus as:

$$e:=\lim_{n\to\infty}\left(1+\frac1n\right)^n$$

My question is this: As $n$ heads to infinity, $1/n$ gets closer and closer to zero. So what we're really doing is to add $1$ to a number that's getting closer and closer to zero and then raising that expression to an increasingly large number.

So, the limit as $n$ moves towards infinity is $1$. Why is it $e$?

Masacroso
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learningMachine
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    A number close to $1$ raised to a big power can be a big number. – Simply Beautiful Art Jan 10 '17 at 12:29
  • Your logic is flawed, as you can see by numerical experimentation. – Matthew Leingang Jan 10 '17 at 12:30
  • yes @SimpleArt ... but the number being added to 1 in this case is an increasingly small number. – learningMachine Jan 10 '17 at 12:30
  • Why do you ignore it's power? – Nosrati Jan 10 '17 at 12:30
  • The thing is that $n$ never really gets to infinity. That's the concept of limit. And because of math it turns out that something like $(1+1/300)^{300}$ is a number near $e$ and that's all. But I think there are many more good explanation on this forum and on the Internet. – Zubzub Jan 10 '17 at 12:31
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    See $(1+\frac1n)^n=1+n\times\frac1n+\cdots=2+\cdots$ with binomial expansion, so It's at least 2 – Nosrati Jan 10 '17 at 12:32
  • The downvotes are idiotic. This is a natural question upon first encountering this. – The Count Jan 10 '17 at 13:15
  • @MyGlasses If only I had thought such a simple thing sooner :D – Simply Beautiful Art Jan 10 '17 at 13:17
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    @TheCount I only downvoted because I feel like Google could answer this, and I also feel as though I've seen similar questions multiple times recently. – Simply Beautiful Art Jan 10 '17 at 13:19
  • @SimpleArt One of my big concerns on this site is that people don't know how to even ask the questions they have, yet we seem to think they should have the background knowledge that we do as math enthusiasts/professionals. So our reactions can turn them off even more. But there is room for disagreement. You certainly have far more experience here than I do. – The Count Jan 10 '17 at 14:38
  • @TheCount ah yes, I know exactly what you are saying, but in my own experience, what you call "turn them off", I call promoting better questions. It is why questions I post have progressively improved and why this site continues to deny homework questions, and I think it's going well. – Simply Beautiful Art Jan 10 '17 at 14:49
  • @SimpleArt fair enough. i do tend to be a bleeding heart, after all. :) – The Count Jan 10 '17 at 14:50
  • @TheCount Please be reminded that, contrarily to a widely held misconception, votes on a given question are not directed at the human being or bot or whichever... thing that is asking the question. Votes on questions primarily exist to segregate the questions of value to the site from the others. Is a question already asked multiple times of value to the site? Ergo. – Did Jan 10 '17 at 15:07
  • @Did Yes, I know that, but again, new users are not familiar with these subtleties. – The Count Jan 10 '17 at 16:08
  • @TheCount And this is your reason for calling "idiotic" some accurate downvotes, thus promoting said misconception? Does not make much sense. – Did Jan 10 '17 at 22:17
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    @Did I am new to the site (and believe one should always strive to be nice) but it seems to me that downvoting such a question is not accurate behaviour. I have just earned the downvoting privilege myself and I read that you should downvote "an egregiously sloppy, no-effort-expended post, or an answer that is clearly and perhaps dangerously incorrect." This does not appear to be the case. Duplicate/off-topic questions appear to involve another mechanism: flagging for moderator attention, or (if one has enough reputation) casting a vote for the question to be closed. Am I wrong? – Anonymous Jan 10 '17 at 23:30
  • @Did You are free to disagree, as stated before. Best. – The Count Jan 11 '17 at 00:42

4 Answers4

4

Here is a table of the sequence $\left(1 + \frac{1}{n}\right)^n$ from Wolfram Alpha:

table

You can see the sequence starts at $2$ and increases. So its limit cannot be $1$.

You write:

So what we're really doing is to add 1 to a number that's getting closer and closer to zero and then raising that expression to an increasingly large number.

This is basically what you're saying: $$ \color{red}{ \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n = \lim_{n\to\infty} 1^n = 1} \qquad\text{(false)} $$ But in this argument, you split the $n$ in the base and the $n$ in the exponent, and took the limit as they tended to $\infty$ separately. But that's not allowed; it's the same variable in both, so we can only take the limit once.

Matthew Leingang
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Let $N$ be a fixed integer. Then as $n$ goes to infinity

  1. on one hand, $$\left(1+\frac1n\right)^N\to1$$
  2. and on the other $$\left(1+\frac1N\right)^n\to\infty.$$

You are focusing on the "1." side, but the "2." side gives a completely different conclusion. In fact, there is no fixed $N$ and all you can tentatively infer for now is that the limit lies somewhere between $1$ and $\infty$ (if it exists).

Using the binomial expansion of $\left(1+\dfrac1n\right)^n$, one can show that the actual limit is given by the converging series

$$e:=\sum_{k=0}^\infty\frac1{k!}.$$

2

I though similarly, but the reason is because even though you think you are adding just $1$, this is not the case. It is true that the second term $\text{approaches 0}$, but it never quite gets there. Thus you always have $1 + \text{some small number}$, raised to exponent, and thus the limit goes to $e$ if you look at Matthew's chart above. It is the $\frac{1}{n}$ that curves the graph upwards, because you are still $\text{adding}$ by a small amount. Although this amount gets less and less and $n \to \infty$, it never disappears.

K Split X
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As @MyGlasses comments, a quick binomial expansion reveals that

$$\left(1+\frac1x\right)^x=1+1+\frac12\frac{x-1}x+\dots\\=2+\frac12\frac{x-1}x+\dots\\>2$$

And from experience, you should realize that all terms in the binomial expansion will be positive.

Simply Beautiful Art
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