# How to show that $\lim_{n\to\infty} \frac{(n+1)^{n-1}}{n^n} =0$?

I know the answer. It's $0$. But how would I get there?

$$\lim_{n\to\infty} \frac{(n+1)^{n-1}}{n^n}$$

$$\lim_{n\to\infty}\frac{(n+1)^{n-1}}{n^n}=\lim_{n\to\infty}\frac1{n+1}\left(\frac{n+1}{n}\right)^n=\lim_{n\to\infty}\frac1{n+1}\cdot\lim_{n\to\infty}\left(1+\frac1n\right)^n=0\cdot e = 0.$$

Hint: use the binomial expansion to get the first few terms (the ones with the highest exponents) of the numerator. Then take the limit of each of the terms.

Note $\frac{(n+1)^{n-1}}{n^n} = \frac{1}{n+1} \frac{(n+1)^n}{n^n}= \frac{1}{n+1} (\frac{n+1}{n})^n = \frac{1}{n+1} (1+\frac{1}{n})^n$. Now, $\frac{1}{n+1} \to 0$ as $n \to \infty$ and $(1+\frac{1}{n})^n \to e$ as $n \to \infty$. So the product goes to $0 \cdot e =0$ as $n \to \infty$.

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

LEt

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

Put $b_n = \frac{1}{n+1}$ and

$c_n = \left( \frac{n+1}{n}\right)^n$.

Clearly, $b_n \to 0$ and $c_n \to e$. Hence

$$\lim a_n = (\lim b_n) ( \lim c_n ) = 0 \times e = 0$$

$$\frac{(n+1)^{n-1}}{n^n}=\frac{(n+1)^{n-1}}{n^{n-1}n}=\left(1+\frac{1}{n} \right)^{n-1}\frac{1}{n}=\left(1+\frac{1}{n} \right)^{n\frac{n-1}{n}}\frac{1}{n} \longrightarrow e^1\cdot 0=0$$ when $n\rightarrow \infty$.