Show $\frac{1}{n - 1} \geq e^\frac{1}{n} - 1, for~n \gt 1$

Question

I am to show that $\frac{1}{n - 1} \geq e^\frac{1}{n} - 1, for~n \in \mathbb{N}^+, n \gt 1$.

I tried substituting using $e^x = \sum_{i=0}^{\infty} \frac{x^i}{i!}$, which gives: $\frac{1}{n-1} \geq \frac{1}{n} + \frac{1}{2n^2} + \frac{1}{6n^3} ...$, and I also tried using the inequality $e^\frac{1}{n-1} \geq e^\frac{1}{n}$, but from there on I am stuck. Can you give me a hint?

Answer 1

We have $e^x \ge 1+x$ for all $x$. Let $x = -{1 \over n}$.

Explicitly:

Then $e^{-{1\over n}} \ge 1-{1 \over n} = {n-1 \over n}$, and so ${ n \over n-1} = 1+ {1 \over n-1} \ge e^{1 \over n}$, or ${1 \over n-1} \ge e^{1 \over n}-1$. This is true for any $n \in \mathbb{R} \setminus \{0,1\}$.

Answer 2

You have said $e^{1/n} - 1 = \frac{1}{n} + \frac{1}{2n^2} + \frac{1}{6n^3} + \ldots$

You could also say $\frac{1}{n-1}= \frac{1}{n} + \frac{1}{n^2} + \frac{1}{n^3} + \ldots$, and then $n\gt 1$ will give you what you want since each term in the second series is greater than or equal to the corresponding term in the first series and both series are absolutely convergent.

Answer 3

A diferente approach. I will show $$ e^x\le\frac{1}{1-x},\quad 0\le x<1. $$ This is equivalent to $e^x(1-x)\le1$. At $x=0$ we have equality. Now $$ (e^x(1-x))'=e^x(1-x)-e^x=-x\,e^x\le0. $$ Thus, $e^x(1-x)$ is decreasing, proving the result.