# show-that-U-n-0-1-dx-1-x-x-2-x-n-converges-

Question Number 131590 by pticantor last updated on 06/Feb/21
$$\boldsymbol{{show}}\:\boldsymbol{{that}} \\$$$$\boldsymbol{{U}}_{\boldsymbol{{n}}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{{dx}}}{\mathrm{1}+\boldsymbol{{x}}+\boldsymbol{{x}}^{\mathrm{2}} +….+\boldsymbol{{x}}^{\boldsymbol{{n}}} }\: \\$$$$\boldsymbol{{converges}}!! \\$$
Answered by mathmax by abdo last updated on 06/Feb/21
$$\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{1}−\mathrm{x}^{\mathrm{n}+\mathrm{1}} }\mathrm{dx}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}−\mathrm{x}^{\mathrm{n}+\mathrm{1}} }\mathrm{dx}\: \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx}\:=\left[\mathrm{x}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{so}\:\mathrm{U}_{\mathrm{n}} \mathrm{cv}\:\mathrm{and}\:\mathrm{U}_{\mathrm{n}} \rightarrow\frac{\mathrm{1}}{\mathrm{2}} \\$$
Commented by pticantor last updated on 08/Feb/21
$$\boldsymbol{{i}}\:\boldsymbol{{think}}\:\boldsymbol{{that}}\:\boldsymbol{{should}}\:\boldsymbol{{be}}\:\boldsymbol{{true}}\:\boldsymbol{{if}} \\$$$$\mid\boldsymbol{{x}}\mid<\mathrm{1} \\$$$$\boldsymbol{{i}}\:\boldsymbol{{have}}\:\boldsymbol{{a}}\:\boldsymbol{{question}}!\:\boldsymbol{{please}}\:\boldsymbol{{in}}\:\boldsymbol{{wich}}\:\boldsymbol{{cases}} \\$$$$\boldsymbol{{we}}\:\boldsymbol{{have}} \\$$$$\boldsymbol{{li}}\underset{\boldsymbol{{n}}>\infty} {\boldsymbol{{m}}}\int_{\boldsymbol{{a}}} ^{\boldsymbol{{b}}} \boldsymbol{{f}}_{\boldsymbol{{n}}} \left(\boldsymbol{{x}}\right)\boldsymbol{{dx}}=\int_{\boldsymbol{{a}}} ^{\boldsymbol{{b}}} \boldsymbol{{li}}\underset{\boldsymbol{{n}}>\infty} {\boldsymbol{{m}f}}_{\boldsymbol{{n}}} \left(\boldsymbol{{x}}\right)\boldsymbol{{dx}}\:? \\$$
Answered by Dwaipayan Shikari last updated on 06/Feb/21
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}}{\mathrm{1}−{x}^{{n}+\mathrm{1}} }{dx}\:\:\:\:\:\:\:\:{x}^{{n}+\mathrm{1}} ={u} \\$$$$\Rightarrow\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{−{n}} −{x}^{\mathrm{1}−{n}} }{\mathrm{1}−{u}}{du}=\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{−\frac{{n}}{{n}+\mathrm{1}}} −{u}^{\frac{\mathrm{1}−{n}}{{n}+\mathrm{1}}} }{\mathrm{1}−{u}}{du} \\$$$$=\frac{\mathrm{1}}{{n}+\mathrm{1}}\left(\psi\left(\mathrm{1}+\frac{\mathrm{1}−{n}}{{n}+\mathrm{1}}\right)−\psi\left(\mathrm{1}−\frac{{n}}{{n}+\mathrm{1}}\right)\right) \\$$$$=\frac{\mathrm{1}}{{n}+\mathrm{1}}\left(\psi\left(\frac{\mathrm{2}}{{n}+\mathrm{1}}\right)−\psi\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)\right) \\$$