Question Number 30175 by abdo imad last updated on 17/Feb/18
$${prove}\:{that}\:{u}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{n}+{k}}\:{is}\:{convergente}\:. \\ $$
Commented by abdo imad last updated on 21/Feb/18
![we have u_n = (1/n) Σ_(k=1) ^n (1/(1 +(k/n))) ⇒ lim_(n→∞) u_n =lim_(n→∞) ((1−0)/n)Σ_(k=1) ^n (1/(1+((k(1−0))/n))) (Rieman sum) = ∫_0 ^1 (dx/(1+x)) =[ln∣1+x∣]_0 ^1 = ln2 .](https://www.tinkutara.com/question/Q30347.png)
$${we}\:{have}\:{u}_{{n}} =\:\frac{\mathrm{1}}{{n}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{1}\:+\frac{{k}}{{n}}}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow\infty} {u}_{{n}} ={lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}−\mathrm{0}}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{{k}\left(\mathrm{1}−\mathrm{0}\right)}{{n}}}\:\:\:\left({Rieman}\:{sum}\right) \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\:\frac{{dx}}{\mathrm{1}+{x}}\:=\left[{ln}\mid\mathrm{1}+{x}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} =\:{ln}\mathrm{2}\:\:. \\ $$
Commented by abdo imad last updated on 21/Feb/18
$${another}\:{method}\:{we}\:{have}\: \\ $$$${u}_{{n}} =\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:+\frac{\mathrm{1}}{{n}+\mathrm{2}}\:+….\:\frac{\mathrm{1}}{\mathrm{2}{n}}=\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+…+\frac{\mathrm{1}}{{n}}\:+\frac{\mathrm{1}}{{n}+\mathrm{1}}\:+… \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}{n}}\:−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+…+\frac{\mathrm{1}}{{n}}\right)={H}_{\mathrm{2}{n}} \:−\:{H}_{{n}} \:{but} \\ $$$${H}_{\mathrm{2}{n}} ={ln}\left(\mathrm{2}{n}\right)\:+\gamma\:\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:{and}\:{H}_{{n}} =\:{ln}\left({n}\right)\:+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\Rightarrow \\ $$$${H}_{\mathrm{2}{n}} \:−{H}_{{n}} ={ln}\left(\frac{\mathrm{2}{n}}{{n}}\right)\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow{lim}_{{n}\rightarrow\infty} {H}_{\mathrm{2}{n}} \:−{H}_{{n}} ={ln}\left(\mathrm{2}\right).{so} \\ $$$${lim}_{{n}\rightarrow\infty} {u}_{{n}} ={ln}\left(\mathrm{2}\right)\:. \\ $$