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If $n \in \mathbb N$ and $n \geq 2$, then we have $\frac 12+ \frac 13+\dots + \frac 1n < \log n < 1 + \frac 12+ \dots + \frac {1}{n-1} $.

My try : Once if we can prove that for all $k \in \mathbb N$ and $k \geq 2$, $\frac 1k < \int_{k-1}^{k} \frac 1t dt < \frac {1}{k-1}$ then we are done.

But I am finding difficulty in proving the inequality $\frac 1k < \int_{k-1}^{k} \frac 1t dt < \frac {1}{k-1}$. Help needed!

Zev Chonoles
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User8976
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1 Answers1

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We just need to prove that for any $n\geq 2$ we have:

$$\frac{1}{n+1}\leq \log\left(1+\frac{1}{n}\right) \leq \frac{1}{n} \tag{1}$$ then apply induction, but $(1)$ is trivial, since $f(x)=\frac{1}{x}$ is a decreasing function on $\mathbb{R}^+$, hence: $$\frac{1}{n+1}\leq\int_{n}^{n+1}\frac{dx}{x}\leq\frac{1}{n}.\tag{2}$$

Jack D'Aurizio
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