(1/(k+1))≤∫_k ^(k+1) ((1/x))dx≤(1/k) please show it with k∈ℵ−(0)


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Question Number 159723 by stelor last updated on 20/Nov/21

$\frac{1}{k + 1} ⩽ \int_{k}^{k + 1} (\frac{1}{x}) d x ⩽ \frac{1}{k}$ $p l e a s e s h o w i t w i t h k \in ℵ - (0)$
Commented by stelor last updated on 20/Nov/21

$\frac{1}{k + 1} ⩽ \int_{k}^{k + 1} (\frac{1}{x}) d x ⩽ \frac{1}{k}$ $p l e a s e s h o w i t w i t h k \in ℵ - (0)$
	Answered by puissant last updated on 20/Nov/21

	$k ⩽ x ⩽ k + 1 \Rightarrow \frac{1}{k + 1} ⩽ \frac{1}{x} ⩽ \frac{1}{k}$ $\Rightarrow \frac{1}{k + 1} \int_{k}^{k + 1} d x ⩽ \int_{k}^{k + 1} \frac{1}{x} d x ⩽ \frac{1}{k} \int_{k}^{k + 1} d x$ $\Rightarrow \frac{1}{k + 1} ⩽ \int_{k}^{k + 1} \frac{1}{x} d x ⩽ \frac{1}{k} . . .$
	Commented by stelor last updated on 20/Nov/21

	$n i c e .$
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