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Is it possible to show using only elementary facts that:

$ \sum_{k=0}^n \frac{1}{k!} \le \left(1 + \frac{1}{2n}\right)^{2n+1} $

Of course they both have the same limit, $e$, but let's assume I don't know that about series.

I guess that I have to use the fact that $\left(1 + \frac{1}{2n}\right)^{2n+1} = \sum_{k=0}^{2n+1} \binom{2n+1}{k} \cdot \frac{1}{(2n)^k}$?

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    The whole task I'm trying to achieve is to show that $\sum_{n=0}^{+\infty} \frac{1}{k!} = e $ using two inequalities. It is easy to find something smaller that has the limit $e$, but I can't find the upper bound that would have the same limit.2012-12-09

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Maybe this does not meet your "elementary" standard, but here goes. It follows directly from the definition of $\log$ $\log(x) =\int_1^x \frac{dt}{t}$ that $1+x \leq e^x$ for all $x\in \mathbb{R}$ and $\tfrac{\partial}{\partial x}e^x = e^x.$

Repeated integration $f \mapsto 1+\int_0^xf$ shows $\sum_{k=0}^n\frac{x^k}{k!} \leq e^x$ for all integral $n\geq 0$ and $x > 0$. On the other hand if $1-x >0$ then after substituting $x \leftarrow -x$ and taking reciprocals it follows that $e^x \leq \frac{1}{1-x}$ and therefore $e^x = e^{(x/n)n} \leq \left(1-\frac{x}{n}\right)^{-n} = \left(1+\frac{x}{n-x}\right)^n$ for all integral $n \geq 1$ and $x < n$. Taking $x = 1$ gives the required (and even a sharper) result.

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    After all, it's the simplest I got here, so I'm going to accept it.2012-12-20