Question Number 126776 by snipers237 last updated on 24/Dec/20
$$\:{Nature}\:{and}\:{Sum}\:\Sigma{u}_{{n}} \\ $$$$\:\:{where}\:{u}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}+\mathrm{1}} }{\mathrm{1}+{x}+…+{x}^{{n}} }{dx} \\ $$
Answered by mindispower last updated on 24/Dec/20
![1+x+......+x^n ≥(n+1).(x^(Σ_(k=0) ^n k) )^(1/(n+1)) =(n+1)x^((n(n+1))/(2(n+1))) ....AM G 1+x+......+x^n ≥(n+1).(x^(Σ_(k=0) ^n k) )^(1/(n+1)) =(n+1)x^((n(n+1))/(2(n+1))) ....AM GM inquality =(1+n)x^(n/2) ⇒U_n ≤∫_0 ^1 (x^((n+2)/2) /(n+1))dx=(2/((n+1)(n+4))) ΣU_n ≤2Σ(1/((n+1)(n+4)))...witch cv 2nd Way starte withe Ψ(z+1)=−γ+∫_0 ^1 ((1−x^z )/(1−x))dx....(I) U_n =∫_0 ^1 (x^(n+1) /(Σ_(k=0) ^n x^k ))dx=∫_0 ^1 ((x^(n+1) (1−x))/(1−x^(n+1) ))dx y=x^(n+1) ⇒dx=(y^(−(n/(n+1))) /(n+1))dy ⇔U_n =(1/(n+1))∫_0 ^1 ((y(1−y^(1/(n+1)) )y^(−(n/(n+1))) )/(1−y))dy U_n =(1/(n+1))∫_0 ^1 ((y^(1/(n+1)) −y^(2/(n+1)) )/(1−y))dy =(1/(n+1))[∫_0 ^1 ((y^(1/(n+1)) −1)/(1−y))dy−∫_0 ^1 ((y^(2/(n+1)) −1)/(1−y))dy] By apply I U_n =((Ψ(((n+3)/(n+1)))−Ψ(((n+2)/(n+1))))/(n+1))=((Ψ(1+(2/(n+1)))−Ψ(1+(1/(n+1))))/(n+1)) Ψ(1+x)−Ψ(1+y)=(y−x)Ψ^1 (1+c) By apply mean value theorem ∃c∈[y,x] ⇒Ψ(1+(2/(n+1)))−Ψ(1+(1/(n+1)))=(1/(n+1))Ψ^1 (1+c) c∈[(1/(n+1));(2/(n+1))] Ψ^1 (z)=Σ_(n≥0) (1/((n+z)^2 )) Ψ^((1)) (1+c)=Σ_(n≥0) (1/((1+c+n)^2 ))≤Σ_(n≥0) (1/((1+n)^2 ))=ζ(2)=(π^2 /6) ⇒U_n =((Ψ(1+(2/(n+1)))−Ψ(1+(1/(n+1))))/(n+1)) Ψ(1+(2/(n+1)))−Ψ(1+(1/(n+1)))≤(1/(n+1)).Ψ^((1)) (1)=(π^2 /(6(n+1))) ⇒U_n ≤(π^2 /(6(n+1)^2 ))⇒ΣU_n ≤ζ^2 (2)=(π^4 /(36))](https://www.tinkutara.com/question/Q126792.png)
$$\mathrm{1}+{x}+……+{x}^{{n}} \geqslant\left({n}+\mathrm{1}\right).\left({x}^{\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{k}} \right)^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} =\left({n}+\mathrm{1}\right){x}^{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}\left({n}+\mathrm{1}\right)}} ….{AM}\:{G} \\ $$$$\mathrm{1}+{x}+……+{x}^{{n}} \geqslant\left({n}+\mathrm{1}\right).\left({x}^{\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{k}} \right)^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} =\left({n}+\mathrm{1}\right){x}^{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}\left({n}+\mathrm{1}\right)}} ….{AM}\:{GM} \\ $$$${inquality} \\ $$$$=\left(\mathrm{1}+{n}\right){x}^{\frac{{n}}{\mathrm{2}}} \\ $$$$\Rightarrow{U}_{{n}} \leqslant\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\frac{{n}+\mathrm{2}}{\mathrm{2}}} }{{n}+\mathrm{1}}{dx}=\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{4}\right)} \\ $$$$\Sigma{U}_{{n}} \leqslant\mathrm{2}\Sigma\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{4}\right)}…{witch}\:{cv} \\ $$$$\mathrm{2}{nd}\:{Way} \\ $$$${starte}\:{withe} \\ $$$$\Psi\left({z}+\mathrm{1}\right)=−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{{z}} }{\mathrm{1}−{x}}{dx}….\left({I}\right) \\ $$$${U}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}+\mathrm{1}} }{\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{x}^{{k}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}+\mathrm{1}} \left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}^{{n}+\mathrm{1}} }{dx} \\ $$$${y}={x}^{{n}+\mathrm{1}} \Rightarrow{dx}=\frac{{y}^{−\frac{{n}}{{n}+\mathrm{1}}} }{{n}+\mathrm{1}}{dy} \\ $$$$\Leftrightarrow{U}_{{n}} =\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{y}\left(\mathrm{1}−{y}^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} \right){y}^{−\frac{{n}}{{n}+\mathrm{1}}} }{\mathrm{1}−{y}}{dy} \\ $$$${U}_{{n}} =\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{y}^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} −{y}^{\frac{\mathrm{2}}{{n}+\mathrm{1}}} }{\mathrm{1}−{y}}{dy} \\ $$$$=\frac{\mathrm{1}}{{n}+\mathrm{1}}\left[\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{y}^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} −\mathrm{1}}{\mathrm{1}−{y}}{dy}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{y}^{\frac{\mathrm{2}}{{n}+\mathrm{1}}} −\mathrm{1}}{\mathrm{1}−{y}}{dy}\right] \\ $$$${By}\:{apply}\:{I} \\ $$$${U}_{{n}} =\frac{\Psi\left(\frac{{n}+\mathrm{3}}{{n}+\mathrm{1}}\right)−\Psi\left(\frac{{n}+\mathrm{2}}{{n}+\mathrm{1}}\right)}{{n}+\mathrm{1}}=\frac{\Psi\left(\mathrm{1}+\frac{\mathrm{2}}{{n}+\mathrm{1}}\right)−\Psi\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)}{{n}+\mathrm{1}} \\ $$$$\Psi\left(\mathrm{1}+{x}\right)−\Psi\left(\mathrm{1}+{y}\right)=\left({y}−{x}\right)\Psi^{\mathrm{1}} \left(\mathrm{1}+{c}\right) \\ $$$${By}\:{apply}\:{mean}\:{value}\: \\ $$$${theorem} \\ $$$$\exists{c}\in\left[{y},{x}\right] \\ $$$$\Rightarrow\Psi\left(\mathrm{1}+\frac{\mathrm{2}}{{n}+\mathrm{1}}\right)−\Psi\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)=\frac{\mathrm{1}}{{n}+\mathrm{1}}\Psi^{\mathrm{1}} \left(\mathrm{1}+{c}\right) \\ $$$${c}\in\left[\frac{\mathrm{1}}{{n}+\mathrm{1}};\frac{\mathrm{2}}{{n}+\mathrm{1}}\right] \\ $$$$\Psi^{\mathrm{1}} \left({z}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({n}+{z}\right)^{\mathrm{2}} } \\ $$$$\Psi^{\left(\mathrm{1}\right)} \left(\mathrm{1}+{c}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{1}+{c}+{n}\right)^{\mathrm{2}} }\leqslant\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{1}+{n}\right)^{\mathrm{2}} }=\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Rightarrow{U}_{{n}} =\frac{\Psi\left(\mathrm{1}+\frac{\mathrm{2}}{{n}+\mathrm{1}}\right)−\Psi\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)}{{n}+\mathrm{1}} \\ $$$$\Psi\left(\mathrm{1}+\frac{\mathrm{2}}{{n}+\mathrm{1}}\right)−\Psi\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)\leqslant\frac{\mathrm{1}}{{n}+\mathrm{1}}.\Psi^{\left(\mathrm{1}\right)} \left(\mathrm{1}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}\left({n}+\mathrm{1}\right)} \\ $$$$\Rightarrow{U}_{{n}} \leqslant\frac{\pi^{\mathrm{2}} }{\mathrm{6}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\Rightarrow\Sigma{U}_{{n}} \leqslant\zeta^{\mathrm{2}} \left(\mathrm{2}\right)=\frac{\pi^{\mathrm{4}} }{\mathrm{36}} \\ $$