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Question Number 182794 by mnjuly1970 last updated on 14/Dec/22

$$$$$$\mathrm{\Omega }=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-\underset{}{\stackrel{}{1}})}^n}{2^n.n\left(n\underset{}{\stackrel{}{+}}1\right)\left(n\underset{}{\stackrel{}{+}}2\right)}=?$$$$$$

Answered by ARUNG_Brandon_MBU last updated on 14/Dec/22

$$S\left(x\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^{n+2}}{n(n+1)(n+2)}\Rightarrow S{}^{\prime }\left(x\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^{n+1}}{n(n+1)}$$$$\Rightarrow S{}^{″}\left(x\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n}\Rightarrow S{}^{‴}\left(x\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}{x}^{n-1}=\frac{1}{1-x}$$$$\Rightarrow S{}^{″}\left(x\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n}=-\ln (1-x)+C_1,C_1=0$$$$\Rightarrow S{}^{\prime }\left(x\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^{n+1}}{n(n+1)}=(1-x)\ln (1-x)+x-1+C_2,C_2=1$$$$\Rightarrow S{}^{\prime }\left(x\right)=\ln (1-x)-x\ln (1-x)+x$$$$\Rightarrow S\left(x\right)=\int [-x\ln (1-x)+\ln (1-x)+x]dx$$$$\Rightarrow S\left(x\right)=-\frac{x^2}{2}\ln (1-x)+\frac{1}{2}(\frac{x^2}{2}+x+\ln \mid x-1\mid )$$$$+(1-x)-(1-x)\ln (1-x)+\frac{x^2}{2}+C_3,C_3=-1$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{2^{n}n(n+1)(n+2)}=2^{2}S(-\frac{1}{2})$$$$=-\frac{1}{2}\ln \left(\frac{3}{2}\right)+2(\frac{1}{8}-\frac{1}{2}+\ln \left(\frac{3}{2}\right))+6-6\ln \left(\frac{3}{2}\right)+\frac{1}{2}-4$$$$\Rightarrow \mathrm{\Omega }=\frac{7}{4}-\frac{9}{2}\ln \left(\frac{3}{2}\right)$$

Commented by mnjuly1970 last updated on 14/Dec/22

$$grateful.thankyousir$$

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