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The question is to prove the inequality $$\frac{1}{n+1} \leq \ln \left(1+\frac 1n\right) \leq \frac 1n\\\forall n \geq 1, n\in \mathbb N$$

I tried using Taylor expansion but couldn't figure out anything. Any ideas? Thanks.

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2 Answers 2

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Suppose $n<x<n+1$. Then $$ \frac{1}{n+1}<\frac{1}{x}<\frac{1}{n}. $$ Integrate this with respect to $x$, from $n$ to $n+1$. Then $$ \frac{1}{n+1}<\log{(n+1)}-\log{n}<\frac{1}{n}, $$ and the middle is $\log{(1+\frac{1}{n})}$.

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  • $\begingroup$ I love how the integration doesn't affect the bounds here. Quite interesting! +1 for sure! $\endgroup$ May 6, 2017 at 0:35
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Given an integer $n>0$, Mean Value Theorem implies that there exist a real number $\xi\in\left(1,1+\frac1n\right)$ s.t. $$\frac{\ln\left(1+\frac1n\right)-\ln1}{\left(1+\frac1n\right)-1}=\frac1{\xi}\qquad\text{i.e.}\qquad \frac n{n+1}<n\ln\left(1+\tfrac1n\right)<1$$ So $$\frac1{n+1}<\ln\left(1+\tfrac1n\right)<\frac1n$$

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    $\begingroup$ (+1) also nice! $\endgroup$
    – Arnaldo
    May 5, 2017 at 19:08
  • $\begingroup$ I always feel like using the Mean Value Theorem is nuking the fly until I remember that it's actually one of the more fundamental theorems in Calculus. $\endgroup$ May 6, 2017 at 0:40

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