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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?

knub
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3 Answers3

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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.

Henry
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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 $n \in \mathbb{R} \setminus \{0,1\}$.

copper.hat
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  • This is nice, as I used the same inequality for showing $e^{\frac{1}{n}} - 1 \geq \frac{1}{n}$, thus giving you upper and lower bounds of $e^{\frac{1}{n}} - 1$. – knub Oct 25 '15 at 22:02
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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.

Julián Aguirre
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