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How do I can prove that $$e\ :=\ \lim_{n\rightarrow\infty}\left(1 + \frac{1}{n}\right)^n\ =\ \lim_{n\rightarrow\infty}\sum_{j=1}^{n}\frac{1}{j!},$$ without use of derivatives?

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This has already been answered as a special case of this answer ($x=1$). –  robjohn Apr 7 '13 at 3:25

2 Answers 2

up vote 4 down vote accepted

We have the equality $$(1+x)^\alpha=\sum_{k=0}^\infty\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}x^k$$ hence for $x=\frac{1}{n}$ and $\alpha=n$ we have $$(1+\frac{1}{n})^n=\sum_{k=0}^\infty\frac{n(n-1)\cdots(n-k+1)}{k!}\frac{1}{n^k}$$ and then \begin{array} \lim\lim_{n\to\infty}(1+\frac{1}{n})^n&=\lim_{n\to\infty}\sum_{k=0}^\infty\frac{n(n-1)\cdots(n-k+1)}{k!}\frac{1}{n^k}\\&=\sum_{k=0}^\infty\lim_{n\to\infty}\frac{n(n-1)\cdots(n-k+1)}{k!}\frac{1}{n^k}\\&=\sum_{k=0}^\infty\frac{1}{k!}.\end{array} Added let's denote by $$f_k(n)=\frac{n(n-1)\cdots(n-k+1)}{k!}\frac{1}{n^k},$$ and since for all $k$ we have $$0\leq\frac{n(n-1)\cdots(n-k+1)}{n^k}\leq 1$$ then $||f_k||_\infty\leq \frac{1}{k!}$ and then the series $\sum_k f_k(n)$ is uniformy convergent for $n$ since the series $\sum_k||f_k||_\infty$ is convergent and this justify the permutation $\lim\sum=\sum\lim$

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@Aryabhata I added the requested justification. –  Sami Ben Romdhane Apr 7 '13 at 1:25
    
@Aryabhata, to downvote a well worked and nicely presented answer for "need of justification" seems to be pretty harsh: like you're grading the answerer! Why wouldn't the answerer leave some stuff for the questioner to think by himself? Even in there are typos/mistakes I think that downvoting well intended/invested answers is against the spirit of this site. –  DonAntonio Apr 7 '13 at 3:09
    
Of course, +1 for a good answer, BTW. –  DonAntonio Apr 7 '13 at 3:09
    
This misses the objective of these questions, imo: -1 for giving up to vote pressure and for giving a complete answer to a homework tagged question. –  DonAntonio Apr 7 '13 at 10:11

You can use the Binomial Series for this. Applying the definition gives:

$$\left(1+\frac{1}{x}\right)^x = \sum_{k \ge 0} \left(\begin{array}{c} x \\ k \end{array}\right)\frac{1}{x^k}$$

where the generalised binomial coefficients are given by:

$$\left( \begin{array}{c} x \\ k \end{array}\right) = \frac{x(x-1)\ldots(x-k+1)}{k!}$$

Putting this together, we have:

$$\left(1+\frac{1}{x}\right)^x = \sum_{k \ge 0} \frac{x(x-1)\ldots(x-k+1)}{k!}\frac{1}{x^k}$$

The key thing here is: what happens as $x \to \infty$? The generalised binomial coefficient is an order $k$ polynomial in $x$. We then divide by $x^k$, so we have something of the form

$$\left(1+\frac{1}{x}\right)^x = \sum_{k \ge 0} \frac{x^k+O(x^{k-1})}{x^k}\frac{1}{k!}$$

Where $O(x^{k-1})$ means something of order $k-1$ in $x$. Hopefully, you can see that

$$\lim_{x \to \infty}\frac{x^k+O(x^{k-1})}{x^k} = 1$$

It follows that:

$$\lim_{x \to \infty}\left(1+\frac{1}{x}\right)^x = \sum_{k \ge 0}\frac{1}{k!}$$

Of course, it doesn't take much more to show that

$$\operatorname{e}^t = \lim_{x \to 0}\left(1+\frac{t}{x}\right)^x = \sum_{k \ge 0} \frac{t^k}{k!}$$

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@JulienClancy I know, I meant the Binomial Series and was correcting it while you were typing your comment. Thanks Julien. –  Fly by Night Apr 6 '13 at 23:35
    
So, for the Binomial Series, I have: –  user70195 Apr 6 '13 at 23:42
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So, for the Binomial Series, I have: $$\left(1 + \frac{1}{n}\right)^n = \sum_{j=0}^n{n \choose j}\frac{1}{n^j} \leq \sum_{j=0}^n\frac{1}{j!},$$ but how do I can obtain the other inequality? Thanx –  user70195 Apr 6 '13 at 23:47
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@Aryabhata The criteria for a down vote is "This answer is not useful". Are you claiming that my answer is not useful? –  Fly by Night Apr 7 '13 at 1:09
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@Aryabhata It is far from misleading; it leads the OP in exactly the right direction. As you say: I just need to apply Abel's Theorem. It is not incorrect because it is the same method as used in one of my calculus books, as another member posted, and as this University website does it: galileo.phys.virginia.edu/classes/152.mf1i.spring02/… –  Fly by Night Apr 7 '13 at 1:15

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