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I have to show that $$\sum_{k=0}^\infty \frac{1}{k!}=e$$ using $$e=\lim_{n\to\infty }(1+\frac{1}{n})^n.$$

My attempt

$$(1+\frac{1}{n})^n=\sum_{k=0}^n\binom{n}{k}\frac{1}{n^k}=\sum_{k=0}^n\frac{n!}{(n-k)!k!}\frac{1}{n^k}.$$ I have that $$\frac{n!}{(n-k)!}=\underbrace{(n-k+1)}_{\leq n}...\underbrace{(n-1)}_{\leq n}\underbrace{n}_{\leq n}\leq n^{k}\implies \frac{n!}{(n-k)!k!}\frac{1}{n^k}\leq \frac{1}{k!},$$ and thus $$(1+\frac{1}{n})^n\leq \sum_{k=0}^n\frac{1}{k!}\implies \lim_{n\to\infty }(1+\frac{1}{n})^n\leq \sum_{k=0}^\infty \frac{1}{k!}\implies e\leq \sum_{k=0}^\infty \frac{1}{k!}.$$ How can I do for the other inequality ?

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  • $\begingroup$ check Theorem 3.31 of Rudin's Principles of Mathematical Analysis. $\endgroup$ – Zhanxiong Oct 27 '15 at 19:12
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Hint: It may be seen as a consequence of the dominated convergence theorem, since for every fixed $k$:

$$ \lim_{n\to +\infty}\frac{n!}{(n-k)!n^k} = \lim_{n\to +\infty}\left(1-\frac{1}{n}\right)\cdot\ldots\cdot\left(1-\frac{k-1}{n}\right) = 1.$$

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  • $\begingroup$ "It may be seen as a consequence of the dominated convergence theorem": I don't see why, but I see how to do solve my exercise thanks to you. Thank you. (but I would like to have a precision about your argument if you have time.) $\endgroup$ – Rick Oct 27 '15 at 20:25
  • $\begingroup$ @Rick: we have to prove that $$\lim_{n\to +\infty}\sum_{k=0}^{n}\frac{1}{k!}\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right) = \lim_{n\to +\infty}\sum_{k=0}^{n}\frac{1}{k!},$$ hence we just have to prove that we are allowed to exchange the limit and the series in the LHS. We are allowed to do so since the LHS is dominated by the RHS, and the RHS is converging. $\endgroup$ – Jack D'Aurizio Oct 27 '15 at 21:02

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