Proving $((n-1)/n)^{n-1}\geq 1/e$

Question

I'm trying to prove that for $n \geq 2$

$\big{(}\frac{n-1}{n}\big{)}^{n-1} \geq \frac{1}{e}$

using induction.

Base case: $n = 2$ $\big{(}\frac{2-1}{2}\big{)}^{2-1} = \frac{1}{2} \geq \frac{1}{e}$

Inductive hypothesis: Assume the inequality holds for some $n = n_0 -1$, so we know

$\big{(}\frac{(n_0-1)-1}{n_0-1}\big{)}^{(n_0-1)-1} \geq \frac{1}{e}$.

We need to prove that

$\big{(}\frac{n_0-1}{n_0}\big{)}^{n_0-1} \geq \frac{1}{e}$.

By hunch is to prove that

$\big{(}\frac{n_0-1}{n_0}\big{)}^{n_0-1} \geq \big{(}\frac{(n_0-1)-1}{n_0-1}\big{)}^{(n_0-1)-1}$

in order to do so. However, I believe we would need to do a lemma to prove that inequality, which makes me think that this proof is getting overly complicated and that I am surely missing something.

Answer 1

Hint Show that $\big{(}\frac{n-1}{n}\big{)}^{n-1}$ is decreasing. What is its limit?

Answer 2

Without induction

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

So, for $n$ sufficiently large, by Taylor $$\log(a_n)=-1+\sum_{p=1}^\infty\frac 1{p(p+1)}\frac 1{n^p}=-1+\frac{1}{2 n}+\frac{1}{6 n^2}+\frac{1}{12n^3}+O\left(\frac{1}{n^4}\right)$$ $$a_n=e^{\log(a_n)}=\frac 1 e \prod_{p=1}^\infty\exp\left(\frac 1{p(p+1)}\frac 1{n^p} \right)$$ $$a_n=e^{\log(a_n)}=\frac 1 e\left(1+\frac{1}{2 n}+\frac{7}{24 n^2}+\frac{3}{16 n^3}+O\left(\frac{1}{n^4}\right)\right)$$