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Question Number 185023 by aba last updated on 17/Jan/23
provet that : Σ_(k=1) ^n (1/(n+k))>ln(((2n+1)/(n+1)))
$$\mathrm{provet}\:\mathrm{that}\::\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}}>\mathrm{ln}\left(\frac{\mathrm{2n}+\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right) \\ $$
Answered by witcher3 last updated on 18/Jan/23
ln(((2n+1)/(n+1)))=∫_(n+1) ^(2n+1) (1/x)dx=Σ_(k=1) ^n ∫_(n+k) ^(k+n+1) (dx/x)≤Σ_(k=1) ^n (1/(n+k))  ∀x∈[n+k,n+k+1];(1/x)≤(1/(n+k))  ⇔Σ_(k=1) ^n (1/(n+k))≥ln(((2n+1)/(n+1)))  >>
$$\mathrm{ln}\left(\frac{\mathrm{2n}+\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right)=\int_{\mathrm{n}+\mathrm{1}} ^{\mathrm{2n}+\mathrm{1}} \frac{\mathrm{1}}{\mathrm{x}}\mathrm{dx}=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\int_{\mathrm{n}+\mathrm{k}} ^{\mathrm{k}+\mathrm{n}+\mathrm{1}} \frac{\mathrm{dx}}{\mathrm{x}}\leqslant\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}} \\ $$$$\forall\mathrm{x}\in\left[\mathrm{n}+\mathrm{k},\mathrm{n}+\mathrm{k}+\mathrm{1}\right];\frac{\mathrm{1}}{\mathrm{x}}\leqslant\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}} \\ $$$$\Leftrightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}}\geqslant\mathrm{ln}\left(\frac{\mathrm{2n}+\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right) \\ $$$$>> \\ $$

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