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Question Number 215315 by depressiveshrek last updated on 02/Jan/25

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

Answered by mr W last updated on 02/Jan/25

$$\underset{n\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^{2n}}\frac{1}{2n+k}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^{2n}}\frac{1}{2n(1+\frac{k}{2n})}$$$$={\int }_0^1\frac{dx}{1+x}$$$$={\left[\ln (1+x)\right]}_0^1$$$$=\ln 2$$

Commented by depressiveshrek last updated on 02/Jan/25

$$\mathrm{Thanks}\mathrm{sir}$$

Answered by MrGaster last updated on 02/Jan/25

$$\underset{n\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^{2n}}\frac{1}{2n+k}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{k=1}{\sum ^{2n}}\frac{1}{2+\frac{k}{n}}$$$$={\int }_0^2\frac{1}{2+x}dx$$$$={[\ln \mid 2+x\mid ]}_0^2$$$$=\ln \mid 4\mid -\ln \mid 2\mid$$$$=\ln \left(\frac{4}{2}\right)$$$$=\ln 2$$

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