Show with help of the Bernoulli Inequality that $\lim_{n\rightarrow\infty}\left(1-\frac{1}{n^2}\right)^{n}=1$ End with: $\lim_{n\rightarrow\infty}\left(1-\frac{1}{n}\right)^n=\frac{1}{e}$
Potence of Euler's Number
2 Answers
Note that Bernoulli's Inequality says that $(1+x)^n \ge 1+ nx$ if $x\ge -1$. In particular, let $x=-\dfrac{1}{n^2}$. We get $1-\frac{n}{n^2} \le \left(1-\frac{1}{n^2}\right)^n \lt 1.$ Now use Squeezing to conclude that the limit is $1$.
Remark: If we have defined $e$ by $e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n,$ then we can use the above result to conclude that since $\left(1-\frac{1}{n^2}\right)^n=\left(1+\frac{1}{n}\right)^n\left(1-\frac{1}{n}\right)^n,$ we must have $\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n=\frac{1}{e}.$
This is a useful step in one approach to defining the exponential function.
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0@RossMillikan: Thanks. The perils of cut and paste! I copied the third to last displayed formula, with intent to modify. Did not modify enough. – 2012-10-17
To ask the second question, if we can use that $ \lim_{n\to\infty}\left(1+\frac{a}{n} \right)=e^a, $ then $ \lim_{n\rightarrow\infty}\left(1-\frac{1}{n^2}\right)^{n}=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n}\left(1+\frac{-1}{n}\right)^{n}=e\cdot e^{-1}=1. $