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Prove that $1+\frac{1}{n} < e$ for all $n$ in the natural numbers. How does this connect to Taylor's formula? I know that $e^x > 1+x$ for $x>0$, but then where does Taylor's formula come in to play?

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  • $\begingroup$ Does it not suffice to show that, because $n$ is a natural number, $\displaystyle \frac{1}{n} \le 1$, and that $\displaystyle 1 + \frac{1}{n} \le 2 \le e$? $\endgroup$ – 2012ssohn Nov 6 '13 at 1:30
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I think what you meant to ask is something along the lines of why $\left(1+\frac{x}{n}\right)^n < e^x$ for positive $x$, since without the exponent the problem is trivial.

In this case, consider the binomial theorem, which is $\sum_{i=0}^n {n \choose i} \left(\frac{x}{n}\right)^i$. Compare this to the Taylor series expansion for $e^x$, which is $\sum_{i=1}^\infty \frac{x^i}{i!}$, and convince yourself that each term of the first summation is less than the corresponding term in the second (provided that $x>0$)

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