Prove that $e^{\frac{1}{n}} \le 1 + \frac{1}{n - 1}$

Asked May 23 '22 at 20:39

Active May 23 '22 at 20:51

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How to prove that:

$$e^{\frac{1}{n}} \le 1 + \frac{1}{n - 1}$$

where $n\in\mathbb{N}, n \ge 2$.

I can't seem to find any connection from description after limit and this equation.

This is part of the following equation proving limit of $\frac{e^{x} - 1}{x} = 1$ for $x \xrightarrow{}0$:

Source: https://www2.karlin.mff.cuni.cz/~halas/MA/MA1/kopacek_-_mat._analyza_nejen_pro_fyziky_1.pdf page 78

edited May 23 '22 at 20:51

asked May 23 '22 at 20:39

meerkat

The proof uses the fact that the sequence $(1+\frac{1}{m})^{m+1}$ approaches $e$ from above, while $(1+\frac{1}{m})^m$ approaches it from below. – Yuval Filmus May 23 '22 at 20:46
The text forms the proof that $\frac{e^x-1}{x} \to 0$. It's not supposed to use it. – Yuval Filmus May 23 '22 at 20:46
@AdamRubinson yes thanks, I've updated condition. – meerkat May 23 '22 at 20:51
@YuvalFilmus I don't get it. Text shows that limit is 1 as x approaches 0 – meerkat May 23 '22 at 20:53
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Right, I had a typo. It should approach 1 instead. – Yuval Filmus May 23 '22 at 20:54
What is the definition of $e$ you are working from? $\sum_k \frac{1}{k!}$ or $\lim_{n\to\infty}\left( 1+1/n\right)^n\ ?$ – Adam Rubinson May 23 '22 at 20:55
@AdamRubinson I am familiar with both approaches, but I guess there will be connection to second one (as it looks similar), just not sure how to connect it to prove this inequality. – meerkat May 23 '22 at 21:03
@AdamRubinson https://www2.karlin.mff.cuni.cz/~halas/MA/MA1/kopacek_-_mat._analyza_nejen_pro_fyziky_1.pdf - page 47 (pdf page 55) – meerkat May 23 '22 at 21:08
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$$e^{\frac{1}{n}}=1+ \frac{1}{n}+\frac{1}{2!}\frac{1}{n^2}+\frac{1}{3!}\frac{1}{n^3}+...$$ $$1 + \frac{1}{n - 1}=\frac{n}{n - 1}=\frac{1}{1 - \frac{1}{n}}=1+\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+...$$ $$e^{\frac{1}{n}}-\Big(1 + \frac{1}{n - 1}\Big)=\Big(\frac{1}{2!}-1\Big)+\Big(\frac{1}{3!}-1\Big)+\Big(\frac{1}{4!}-1\Big)+...<0$$ – Svyatoslav May 23 '22 at 21:34
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Does this answer your question? Show $\frac{1}{n - 1} \geq e^\frac{1}{n} - 1, for~n \gt 1$ . Found using Approach0. There's also the more general case for any real $x \gt 1$ in $e^{\frac{1}{x}} < 1 + \frac{1}{x-1} $. – John Omielan May 23 '22 at 21:46
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This is the Taylor's series $e^x=1+x+\frac{x^2}{2!}+...$ – Svyatoslav May 24 '22 at 16:01

Prove that $e^{\frac{1}{n}} \le 1 + \frac{1}{n - 1}$

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