inequality - How to prove $(1-frac{1}{n})^n leq frac{1}{e} leq(1-frac{1}{n})^{n-1}$?

How to show $(1-\frac{1}{n})^n \leq \frac{1}{e} \leq(1-\frac{1}{n})^{n-1}$?

I can prove the first inequality: take the logarithm of both sides and then use the fact that $\log(1+x) \leq x.$

But how to prove the second inequality? The same method does not work here because we need an lower bound for $\log(1+x)$ now.

Answer

Take logarithms in the second inequality to get $$-1 \le (n-1) \log(1-1/n)$$ which rearranges to $$- \log(1-1/n) \le \frac{1}{n-1}.$$ You can write this as $$\int_{1-1/n}^1 \frac 1t \, dt \le \frac{1}{n-1}.$$

Since $f(t) = \frac 1t$ is decreasing on $[1-1/n,1]$ its maximum value there is $1/(1-1/n) = n/(n-1)$. Consequently $$\int_{1-1/n}^1 \frac 1t \, dt \le \frac{n}{n-1} \cdot \frac 1n = \frac 1{n-1}.$$