Why is$ (1+\frac{1}{n})^n=e$ when n goes to infinity?

Question

Why is $\lim\limits_{n\to\infty}(1+\frac1n)^n=e$?

I think it involves $\sum\limits_{n=0}^\infty\frac1{k!}=e$ but not sure how to get from one to the other.

At first we need how you did define $e$, because often $e$ is defined as this limit — Dominic Michaelis, Aug 25 '13 at 18:18
well before the taylor expansion I knew e as ln(1), the limit is the third definition of e that I encountered. EDIT: which is wrong, oops — Leo, Aug 25 '13 at 18:19
$e$ is not $\ln(1)$. $\ln(1) = 0$. You probably meant $e$ is the unique real number such that $\ln(e) = 1$. But for that to be meaningful it is useful to know how you define the natural log. — Alex Wertheim, Aug 25 '13 at 18:20
Similar question posted and deleted 15 hours ago (by different account): http://math.stackexchange.com/questions/475493/basic-question-on-limit — Jonas Meyer, Aug 25 '13 at 18:23
the answer to my question is indeed in the possible duplicate. However I must admit I dont understand the answer given there. — Leo, Aug 25 '13 at 18:24
e is often defined this way. since you asked why, the answer depends on the definition of limit.Do you know how to define limit?
What you probably are expecting is not that kind of an answer. — jaseem, Aug 25 '13 at 18:24
In the possible duplicate they state $\frac{1}{n^k} {n \choose k} \to \frac{1}{k!}$ as $n \to \infty$, I dont see how this works. Is it in the definition of ${n \choose k} $? For .SE ettiquette: should I delete my question now that it is marked as a duplicate? — Leo, Aug 25 '13 at 18:26
@Leo Search the site. There is also this question with several answers, and most probably many more. — Start wearing purple, Aug 25 '13 at 18:28
You are right. The answer there explains what I ask in my comment above and is exactly what I was hoping to get here. Should I delete this question? — Leo, Aug 25 '13 at 18:30
I agree with Dominic. The more beautiful approach to introduce the ln is through an integral based upon y = 1/x Based upon this assumption and the limit definition of the derivative on y=1/x, you can derive the limit version of the ln. From here you can dervie the derivative of the exponential function from which the e-power arises. I can't explain it here, but if you read a calc book with this introduction of the ln, your question will be answered very clear. — imranfat, Aug 25 '13 at 20:11
Euler original definition of ${\rm e}^{x}$ was to look for a function ${\rm f}\left(x\right)$ which satisfies ${\rm f}'\left(x\right) = {\rm f}\left(x\right), \forall\ x$. Then, ${\rm f}^{\left(n\right)}\left(0\right) = {\rm f}\left(0\right)$. Then, you get ${\rm f}\left(x\right) = {\rm f}\left(0\right)\sum_{n = 1}^{\infty}x^{n}/n!$. — Felix Marin, Sep 05 '13 at 21:37

score 3 · Accepted Answer · edited Aug 25 '13 at 19:58

3

Have you tried expanding by the binomial theorem? ;-) $$(1/n+1)^n=\sum_{k=0}^n\binom{n}k\frac1{n^k}=\sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac1{n^k}$$then as $n\to\infty$ we find $n!/(n^k (n-k)!)\to1$ hence we have:$$\lim_{n\to\infty}\sum_{k=0}^n\frac{n!}{k!(n-k)!\cdot n^k}=\sum_{k=0}^\infty\frac1{k!}\equiv e$$

edited Aug 25 '13 at 19:58

Grobber

3,248

answered Aug 25 '13 at 18:19

obataku

5,571

Why does the limit is $1$ ? – Belgi Aug 25 '13 at 19:35
3

$\displaystyle\frac{n!}{(n-k)!n^k}=\frac{n(n-1)\cdots (n-k+1)}{n^k}\to 1$.$ – Grobber Aug 25 '13 at 19:54
oops -- did not see that earlier. @Belgi $n!/(n-k)!=n(n-1)\cdots(n-k+1)$ and so $n!/((n-k)!n^k)=n/n\cdot(n-1)/n\cdots (n-k+1)/n$ – obataku Aug 25 '13 at 19:56
observe $(n-r)/n=1-r/n$ hence as $n\to\infty$ each of our product terms tends to $1$ – obataku Aug 25 '13 at 19:57
@Grobber - thanks! – Belgi Aug 25 '13 at 20:01

Why is$ (1+\frac{1}{n})^n=e$ when n goes to infinity?

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