Question Number 197660 by pticantor last updated on 25/Sep/23
Answered by witcher3 last updated on 25/Sep/23
$$\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}+...\mathrm{t}^{\mathrm{n}} }\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{dt}=\mathrm{1} \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{dt}}{\mathrm{1}+...+\mathrm{t}^{\mathrm{n}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{t}}{\mathrm{1}−\mathrm{t}^{\mathrm{n}+\mathrm{1}} }\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{1}−\mathrm{tdt}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$