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Question Number 107287 by mathmax by abdo last updated on 09/Aug/20
$$\mathrm{find}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \sum_{\mathrm{k}=\mathrm{n}} ^{\mathrm{2n}−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{k}+\mathrm{n}} \\ $$
Answered by mathmax by abdo last updated on 10/Aug/20
![U_n =Σ_(k=n) ^(2n−1) (1/(k+n)) ⇒U_n =_(k−n =p) Σ_(p=0) ^(n−1) (1/(n+p+n)) =Σ_(p=0) ^(n−1) (1/(2n+p)) =(1/n)Σ_(p=0) ^(n−1) (1/(2+(p/n))) ⇒lim_(n→+∞) U_n =∫_0 ^1 (dx/(2+x)) =[ln(x+2)]_0 ^1 =ln(3)−ln(2)](Q107413.png)
$$\mathrm{U}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{n}} ^{\mathrm{2n}−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{k}+\mathrm{n}}\:\:\Rightarrow\mathrm{U}_{\mathrm{n}} =_{\mathrm{k}−\mathrm{n}\:=\mathrm{p}} \:\:\:\:\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{n}+\mathrm{p}+\mathrm{n}}\:=\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{p}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{p}}{\mathrm{n}}}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\mathrm{2}+\mathrm{x}}\:=\left[\mathrm{ln}\left(\mathrm{x}+\mathrm{2}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{ln}\left(\mathrm{3}\right)−\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$