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Question Number 185024 by aba last updated on 15/Jan/23

$$\mathrm{prove}\mathrm{that}:\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{n}+\mathrm{k}}<\ln \left(2\right)$$

Answered by witcher3 last updated on 16/Jan/23

$$U_n=\underset{k=1}{\sum ^n}\frac{1}{n+k},U_{n+1}-U_n=-\frac{1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}$$ $$=\frac{1}{2(n+1)(2n+1)}>0,U_{n}increase$$ $$U_n⩽\underset{n\to \mathrm{\infty }}{\lim }{U}_n=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{k=1}{\sum ^n}\frac{1}{1+\frac{k}{n}}={\int }_0^1\frac{1}{1+x}dx=ln\left(2\right)$$ $$RiemannSum$$

Commented byaba last updated on 16/Jan/23

$$\mathrm{using}\mathrm{the}\mathrm{relation}:\frac{1}{\mathrm{b}}<\frac{\ln \left(\mathrm{b}\right)-\ln \left(\mathrm{a}\right)}{\mathrm{b}-\mathrm{a}}<\frac{1}{\mathrm{a}}(0<\mathrm{a}<\mathrm{b})$$ $$\mathrm{let}\mathrm{b}=\mathrm{n}+\mathrm{k}\mathrm{and}\mathrm{a}=\mathrm{n}+\mathrm{k}-1$$ $$\frac{1}{\mathrm{n}+\mathrm{k}}<\frac{\ln (\mathrm{n}+\mathrm{k})-\ln (\mathrm{n}+\mathrm{k}-1)}{\mathrm{n}+\mathrm{k}-\mathrm{n}-\mathrm{k}+1}<\frac{1}{\mathrm{n}+\mathrm{k}-1}$$ $$\Rightarrow \frac{1}{\mathrm{n}+\mathrm{k}}<\ln (\mathrm{n}+\mathrm{k})-\ln (\mathrm{n}+\mathrm{k}-1)<\frac{1}{\mathrm{n}+\mathrm{k}-1}$$ $$\Rightarrow \frac{1}{\mathrm{n}+\mathrm{k}}<\ln (\mathrm{n}+\mathrm{k})-\ln (\mathrm{n}+\mathrm{k}-1)<\frac{1}{\mathrm{n}+\mathrm{k}-1}$$ $$\Rightarrow \frac{1}{\mathrm{n}+\mathrm{k}}<\ln (\mathrm{n}+\mathrm{k})-\ln (\mathrm{n}+\mathrm{k}-1)$$ $$\Rightarrow \underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{n}+\mathrm{k}}<\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}(\ln (\mathrm{n}+\mathrm{k})-\ln (\mathrm{n}+\mathrm{k}-1))$$ $$\Rightarrow \underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{n}+\mathrm{k}}<\ln \left(2\mathrm{n}\right)-\ln (2\mathrm{n}-1)+\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}-1}}(\ln (\mathrm{n}+\mathrm{k})-\ln (\mathrm{n}+\mathrm{k}-1))$$ $$\Rightarrow \underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{n}+\mathrm{k}}<\ln \left(2\mathrm{n}\right)-\ln (2\mathrm{n}-1)-\ln \left(\mathrm{n}\right)+\ln (2\mathrm{n}-1)$$ $$\Rightarrow \underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{n}+\mathrm{k}}<\ln \left(2\mathrm{n}\right)-\ln \left(\mathrm{n}\right)$$ $$\Rightarrow \underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{n}+\mathrm{k}}<\ln \left(2\right)$$

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