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Question Number 209926 by depressiveshrek last updated on 26/Jul/24

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{3n+1}+\frac{1}{3n+2}+...+\frac{1}{4n}$$

Commented by Frix last updated on 26/Jul/24

$$\underset{k=1}{\sum ^n}\frac{1}{3n+k}=H_{4n}-H_{3n}$$$$a>b:\underset{n\to \mathrm{\infty }}{\lim }(H_{an}-H_{bn})=\ln \frac{a}{b}$$$$\Rightarrow \mathrm{Answer}\mathrm{is}\ln \frac{4}{3}$$

Commented by depressiveshrek last updated on 26/Jul/24

$$\mathrm{What}\mathrm{is}\mathrm{that}\mathrm{notation}?\mathrm{Can}\mathrm{you}\mathrm{please}$$$$\mathrm{elaborate}\mathrm{further}?$$

Commented by mr W last updated on 26/Jul/24

$$H_n=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$$

Answered by mr W last updated on 26/Jul/24

$$=\underset{n\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^n}\frac{1}{3n+k}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^n}\frac{1}{3+\frac{k}{n}}\times \frac{1}{n}$$$$={\int }_0^1\frac{dx}{3+x}$$$$={\left[\ln (3+x)\right]}_0^1=\ln \frac{4}{3}$$

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