Question Number 209926 by depressiveshrek last updated on 26/Jul/24
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{2}}+…+\frac{\mathrm{1}}{\mathrm{4}{n}} \\ $$
Commented by Frix last updated on 26/Jul/24
$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}{n}+{k}}\:={H}_{\mathrm{4}{n}} −{H}_{\mathrm{3}{n}} \\ $$$${a}>{b}:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left({H}_{{an}} −{H}_{{bn}} \right)\:=\mathrm{ln}\:\frac{{a}}{{b}} \\ $$$$\Rightarrow\:\mathrm{Answer}\:\mathrm{is}\:\:\:\:\:\mathrm{ln}\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$
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
$${H}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{{n}} \\ $$
Answered by mr W last updated on 26/Jul/24
![=lim_(n→∞) Σ_(k=1) ^n (1/(3n+k)) =lim_(n→∞) Σ_(k=1) ^n (1/(3+(k/n)))×(1/n) =∫_0 ^1 (dx/(3+x)) =[ln (3+x)]_0 ^1 =ln (4/3)](https://www.tinkutara.com/question/Q209930.png)
$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{3}{n}+{k}} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{3}+\frac{{k}}{{n}}}×\frac{\mathrm{1}}{{n}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{3}+{x}} \\ $$$$=\left[\mathrm{ln}\:\left(\mathrm{3}+{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{ln}\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$