# Show how $\lim\limits_{n\to\infty}\left(\frac{n-1}{n+1}\right)^n = \frac{1}{e^2}$

How does one evaluate this limit? $$\lim\limits_{n\to\infty}\left(\frac{n-1}{n+1}\right)^n$$

I got to $$\lim_{n\to\infty}\exp\left(n\cdot\ln\left(\frac{n-1}{n+1}\right)\right)$$ but I'm not sure where to go from there.

• Maybe you already know that the limit as $k$ goes to $\infty$ of $(1+x/k)^k$ is $e^x$. Then you can rewrite $\frac{n-1}{n+1}$ as $1-\frac{2}{n+1}$. – André Nicolas Nov 3 '15 at 16:56

Hint: $$\left(\frac{n-1}{n+1}\right)^n=\frac{\left(1-\frac{1}{n}\right)^n}{\left(1+\frac{1}{n}\right)^n}\longrightarrow\frac{e^{-1}}{e}$$

• 25 seconds faster than I! ;-) – Mark Viola Nov 3 '15 at 17:06
• Great minds think alike! – Wojowu Nov 3 '15 at 17:12

$$\lim\limits_{n\to\infty}\left(\frac{n-1}{n+1}\right)^n = \lim\limits_{n\to\infty}\left(1-\frac2{n+1}\right)^n = \lim\limits_{n\to\infty}\left(1-\frac2{n}\right)^n$$ Let $\frac{1}{x} = -\frac{2}{n} \Rightarrow$ $$\lim\limits_{n\to\infty}\left(\frac{n-1}{n+1}\right)^n = \lim\limits_{n\to\infty}\left(1-\frac2{n}\right)^n = \lim\limits_{n\to\infty}\left(1+\frac1{x}\right)^{-2x} = \left( \lim\limits_{n\to\infty}\left(1+\frac1{x}\right) ^ x \right)^{-2} = e ^{-2} = \frac{1}{e^2}$$

One way to proceed is to write

$$\left(\frac{n-1}{n+1}\right)^n=\frac{\left(1-\frac1n\right)^n}{\left(1+\frac1n\right)^n}$$

As $n\to \infty$, the numerator approaches $e^{-1}$ and the denominator approaches $e$.

You could use that $$\frac{n-1}{n+1}=1-\frac2{n+1}\text{ or }=\frac{1-n^{-2}}{(1+n^{-1})^2}.$$

\begin{align}L&=\lim_\limits{n\to\infty}\left(\dfrac{n-1}{n+1}\right)^n\\&=\lim_\limits{n\to\infty}\left(\dfrac{n+1-2}{n+1}\right)^n\\&=\lim_\limits{n\to\infty}\left\{\left(1-\dfrac2{n+1}\right)^{n+1}\right\}^{\frac n{n+1}}\\&=(e^{-2})^{\lim_\limits{n\to\infty}\frac n{n+1}}\\&=e^{-2}\\&=\dfrac1{e^2}\end{align}

$$n\ln\left(\frac{n-1}{n+1}\right)=n\ln\left(1-\frac{2}{n+1}\right)=-2\frac{\ln\left(1-\frac{2}{n+1}\right)}{-\frac{2}{n+1}}-\ln\left(1-\frac{2}{n+1}\right)$$

and use the fact that $\lim_{u\to 0}\frac{\ln(1+u)}{u}=1$ and that $x\mapsto e^x$ is continuous at $x=-2$.

$$\lim_{n\to\infty} \left(\frac{n-1}{n+1}\right)^n=$$ $$\lim_{n\to\infty} \exp\left(\ln\left(\left(\frac{n-1}{n+1}\right)^n\right)\right)=$$ $$\lim_{n\to\infty} \exp\left(n\ln\left(\frac{n-1}{n+1}\right)\right)=$$ $$\exp\left(\lim_{n\to\infty}n\ln\left(\frac{n-1}{n+1}\right)\right)=$$ $$\exp\left(\lim_{n\to\infty}\frac{\ln\left(\frac{n-1}{n+1}\right)}{\frac{1}{n}}\right)=$$ $$\exp\left(\lim_{n\to\infty}\frac{\frac{\text{d}}{\text{d}n}\ln\left(\frac{n-1}{n+1}\right)}{\frac{\text{d}}{\text{d}n}\left(\frac{1}{n}\right)}\right)=$$ $$\exp\left(\lim_{n\to\infty}-\frac{2n^2}{(n-1)(n+1)}\right)=$$ $$\exp\left(\lim_{n\to\infty}-\frac{2n^2}{n(n+1)}\right)=$$ $$\exp\left(\lim_{n\to\infty}-\frac{2}{1+\frac{1}{n}}\right)=$$ $$\exp\left(-\frac{2}{1}\right)=$$ $$\exp\left(-2\right)=e^{-2}=\frac{1}{e^2}$$