0

$$\frac{1}{e}=\lim_{n\to \infty}\left(1-\frac{1}{n+1}\right)^{n+1}$$ How?

To begin I must say that I am aware that: $$e=\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n$$ but this I do not know how to get. Help?

Empty
  • 13,012
Bozo Vulicevic
  • 2,836
  • See http://math.stackexchange.com/questions/1350338/how-to-prove-that-lim-limits-n-to-infty-left-1-frac1n-rightn-fr – peter a g Oct 18 '15 at 15:09
  • in general, you have $\left(1 + small\right)^{LARGE}=e^{small \times LARGE} $ – abel Oct 18 '15 at 15:24

2 Answers2

6

Try to use $$ 1-\frac1{n+1}=\frac{n}{n+1}=\left(1+\frac1n\right)^{-1}. $$

Lutz Lehmann
  • 126,666
3

First notice that $a_n=\Big(1-\frac{1}{n+1}\Big)^{n+1}$ is a subsequence of $b_n = \Big(1-\frac{1}{n}\Big)^{n}$, so it suffices to show that $\lim_{n\rightarrow\infty}b_n = e^{-1}$.

Write $y=\Big(1-\frac{1}{x}\Big)^{x}$. Applying the logarithm to both sides you get

$$\ln y = x\ln\Big(1-\frac{1}{x}\Big)$$

Now the limit $\lim_{x \rightarrow \infty} x\ln\Big(1-\frac{1}{x}\Big)$ is indeterminate $0\cdot\infty$ type, so we can solve it by doing

\begin{eqnarray*} \lim_{x \rightarrow \infty} x\ln\Big(1-\frac{1}{x}\Big) &=& \lim_{x \rightarrow \infty} \frac{\ln\Big(1-\frac{1}{x}\Big)}{x^{-1}} \\ &=&\lim_{x \rightarrow \infty} \frac{\frac{1}{1-1/x}(-1/x)'}{(1/x)'} \\ &=&-\lim_{x \rightarrow \infty} \frac{1}{1-1/x} \\ &=& -1 \end{eqnarray*}

by L'Hopital's rule. By continuity of the logarithm function, we get

$$-1=\lim_{x\rightarrow\infty} \ln y = \ln \Big(\lim_{x\rightarrow\infty}y\Big),$$

so $\lim_{x\rightarrow\infty}\Big(1-\frac{1}{x}\Big)^{x}=\lim_{x\rightarrow\infty} y = e^{-1}$, hence $\lim_{n\rightarrow\infty}b_n = \lim_{n\rightarrow\infty}a_n=\frac{1}{e}$.

Gondim
  • 816