n-0-a-n-n-2-2-n-4-2-n-3-2-

Question Number 195436 by MrGHK last updated on 02/Aug/23

Σ_(n=0) ^∞ (((−a)^(−n) )/((n+2)^2 ))(𝛙(((n+4)/2))−𝛙(((n+3)/2)))=???

$$\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\boldsymbol{\mathrm{a}}\right)^{−\boldsymbol{\mathrm{n}}} }{\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right)^{\mathrm{2}} }\left(\boldsymbol{\psi}\left(\frac{\boldsymbol{\mathrm{n}}+\mathrm{4}}{\mathrm{2}}\right)−\boldsymbol{\psi}\left(\frac{{n}+\mathrm{3}}{\mathrm{2}}\right)\right)=??? \\ $$

Answered by witcher3 last updated on 02/Aug/23

Ψ(1+x)=−γ+∫_0 ^1 ((1−x^s )/(1−x))dx S=Σ_(n≥0) (((−a)^(−n) )/((n+2)^2 ))(𝚿(((n+4)/2))−Ψ(((n+3)/2))) =Σ(((−a)^(−n) )/((n+2)))(∫_0 ^1 ((x^((n+1)/2) −x^((n+2)/2) )/(1−x))dx =a^2 Σ_(n≥0) (((−(1/a))^(n+2) )/((n+2)^2 ))∫_0 ^1 (x^((n+1)/2) /(1+(√x)))dx =a^2 Σ_(n≥0) ((((−(1/a))^(n+2) )/((n+2)^2 ))∫_0 ^1 (x^((n+2)/2) /( (√x)(1+(√x))))dx,(√x)=y,∀∣a∣≥1 =2a^2 ∫_0 ^1 Σ_(n≥0) (((−(y/a))^(n+2) )/((n+2)^2 )).(dy/(1+y)) Σ_(n≥1) (x^n /n^2 )=Li_2 (x) =2a^2 ∫_0 ^1 ((Li_2 (−(y/a))+(y/a))/(1+y))dy =2a^2 ∫_0 ^1 ((Li_2 (−(y/a)))/(1+y))dy+2a∫_0 ^1 (y/(1+y))dy =2a^3 ∫_0 ^(1/a) ((Li_2 (−y))/(1+ay))...verry long calculate

$$\Psi\left(\mathrm{1}+\mathrm{x}\right)=−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\mathrm{x}^{\mathrm{s}} }{\mathrm{1}−\mathrm{x}}\mathrm{dx} \\ $$$$\mathrm{S}=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{a}\right)^{−\boldsymbol{\mathrm{n}}} }{\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right)^{\mathrm{2}} }\left(\boldsymbol{\Psi}\left(\frac{\mathrm{n}+\mathrm{4}}{\mathrm{2}}\right)−\Psi\left(\frac{\mathrm{n}+\mathrm{3}}{\mathrm{2}}\right)\right) \\ $$$$=\Sigma\frac{\left(−\mathrm{a}\right)^{−\mathrm{n}} }{\left(\mathrm{n}+\mathrm{2}\right)}\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}} −\mathrm{x}^{\frac{\mathrm{n}+\mathrm{2}}{\mathrm{2}}} }{\mathrm{1}−\mathrm{x}}\mathrm{dx}\right. \\ $$$$=\mathrm{a}^{\mathrm{2}} \underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\frac{\mathrm{1}}{\mathrm{a}}\right)^{\mathrm{n}+\mathrm{2}} }{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+\sqrt{\mathrm{x}}}\mathrm{dx} \\ $$$$=\mathrm{a}^{\mathrm{2}} \underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(\frac{\left(−\frac{\mathrm{1}}{\mathrm{a}}\right)^{\mathrm{n}+\mathrm{2}} }{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\frac{\mathrm{n}+\mathrm{2}}{\mathrm{2}}} }{\:\sqrt{\mathrm{x}}\left(\mathrm{1}+\sqrt{\mathrm{x}}\right)}\mathrm{dx},\sqrt{\mathrm{x}}=\mathrm{y},\forall\mid\mathrm{a}\mid\geqslant\mathrm{1}\right. \\ $$$$=\mathrm{2a}^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} \underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\frac{\mathrm{y}}{\mathrm{a}}\right)^{\mathrm{n}+\mathrm{2}} }{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }.\frac{\mathrm{dy}}{\mathrm{1}+\mathrm{y}} \\ $$$$\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }=\mathrm{Li}_{\mathrm{2}} \left(\mathrm{x}\right) \\ $$$$=\mathrm{2a}^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{Li}_{\mathrm{2}} \left(−\frac{\mathrm{y}}{\mathrm{a}}\right)+\frac{\mathrm{y}}{\mathrm{a}}}{\mathrm{1}+\mathrm{y}}\mathrm{dy} \\ $$$$=\mathrm{2a}^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{Li}_{\mathrm{2}} \left(−\frac{\mathrm{y}}{\mathrm{a}}\right)}{\mathrm{1}+\mathrm{y}}\mathrm{dy}+\mathrm{2a}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{y}}{\mathrm{1}+\mathrm{y}}\mathrm{dy} \\ $$$$=\mathrm{2a}^{\mathrm{3}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{a}}} \frac{\mathrm{Li}_{\mathrm{2}} \left(−\mathrm{y}\right)}{\mathrm{1}+\mathrm{ay}}…\mathrm{verry}\:\mathrm{long}\:\mathrm{calculate} \\ $$$$ \\ $$

Commented by MathedUp last updated on 02/Aug/23

Wow You R Hero LOL u always come out everywhere :)

$$\mathrm{Wow}\:\mathrm{You}\:\mathrm{R}\:\mathrm{Hero}\:\mathrm{LOL} \\ $$$$\left.\mathrm{u}\:\mathrm{always}\:\mathrm{come}\:\mathrm{out}\:\mathrm{everywhere}\::\right) \\ $$

Commented by MrGHK last updated on 03/Aug/23

Σ_(n=0) ^∞ (((−a)^(−n) )/((n+2)^2 ))(𝛙(((n+4)/2))−𝛙(((n+3)/2))) You are really a brilliant man

Commented by witcher3 last updated on 04/Aug/23

thak You god bless You hard[worck and passion

$$\mathrm{thak}\:\mathrm{You}\:\mathrm{god}\:\mathrm{bless}\:\mathrm{You}\:\mathrm{hard}\left[\mathrm{worck}\:\mathrm{and}\:\mathrm{passion}\right. \\ $$

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