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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
up vote 5 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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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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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
@ShanaKugimiya There's no need for inequalities. I've written a full reply above. All you really need to be able to do is find the limit of a polynomial divided by a polynomial. – Fly by Night Apr 6 '13 at 23:56

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