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Question Number 159723 by stelor last updated on 20/Nov/21
(1/(k+1))≤∫_k ^(k+1) ((1/x))dx≤(1/k)            please show it with k∈ℵ−(0)
$$\frac{\mathrm{1}}{{k}+\mathrm{1}}\leqslant\int_{{k}} ^{{k}+\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right){dx}\leqslant\frac{\mathrm{1}}{{k}}\:\:\:\:\:\:\:\:\: \\ $$$$\:{please}\:{show}\:{it}\:{with}\:{k}\in\aleph−\left(\mathrm{0}\right) \\ $$$$ \\ $$
Commented by stelor last updated on 20/Nov/21
(1/(k+1))≤∫_k ^(k+1) ((1/x))dx≤(1/k)            please show it with k∈ℵ−(0)
$$\frac{\mathrm{1}}{{k}+\mathrm{1}}\leqslant\int_{{k}} ^{{k}+\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right){dx}\leqslant\frac{\mathrm{1}}{{k}}\:\:\:\:\:\:\:\:\: \\ $$$$\:{please}\:{show}\:{it}\:{with}\:{k}\in\aleph−\left(\mathrm{0}\right) \\ $$$$ \\ $$
Answered by puissant last updated on 20/Nov/21
  k ≤ x ≤ k+1   ⇒  (1/(k+1)) ≤ (1/x) ≤ (1/k)   ⇒ (1/(k+1))∫_k ^(k+1) dx ≤ ∫_k ^(k+1) (1/x) dx ≤ (1/k)∫_k ^(k+1) dx  ⇒ (1/(k+1)) ≤ ∫_k ^(k+1) (1/x) dx ≤ (1/k)...
$$\:\:{k}\:\leqslant\:{x}\:\leqslant\:{k}+\mathrm{1}\:\:\:\Rightarrow\:\:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:\leqslant\:\frac{\mathrm{1}}{{x}}\:\leqslant\:\frac{\mathrm{1}}{{k}}\: \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{k}+\mathrm{1}}\int_{{k}} ^{{k}+\mathrm{1}} {dx}\:\leqslant\:\int_{{k}} ^{{k}+\mathrm{1}} \frac{\mathrm{1}}{{x}}\:{dx}\:\leqslant\:\frac{\mathrm{1}}{{k}}\int_{{k}} ^{{k}+\mathrm{1}} {dx} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:\leqslant\:\int_{{k}} ^{{k}+\mathrm{1}} \frac{\mathrm{1}}{{x}}\:{dx}\:\leqslant\:\frac{\mathrm{1}}{{k}}… \\ $$
Commented by stelor last updated on 20/Nov/21
nice.
$${nice}. \\ $$

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