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Question Number 33710 by math khazana by abdo last updated on 22/Apr/18

$$1)findtheradiusofconvergence?for$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n(n+1)(n+2)}andcalculateitssum$$$$$$$$2)findthevalueof\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n{2}^n(n+1)(n+2)}$$

Commented by prof Abdo imad last updated on 27/Apr/18

$$letput{u}_n=\frac{1}{n(n+1)(n+2)}wehave$$$$\frac{u_{n+1}}{u_n}=\frac{n(n+1)(n+2)}{(n+1)(n+2)(n+3)}\Rightarrow {lim}_{n\to +\mathrm{\infty }}\frac{u_{n+1}}{u_n}=1$$$$\Rightarrow R=1letdecomposeF\left(x\right)=\frac{1}{x(x+1)(x+2)}$$$$F\left(x\right)=\frac{a}{x}+\frac{b}{x+1}+\frac{c}{x+2}$$$$a={lim}_{x\to 0}xF\left(x\right)=\frac{1}{2}$$$$b={lim}_{x\to -1}(x+1)F\left(x\right)=\frac{1}{-1}=-1$$$$c={lim}_{x\to -2}(x+2)F\left(x\right)=\frac{1}{(-2)(-1)}=\frac{1}{2}$$$$\Rightarrow F\left(x\right)=\frac{1}{2x}-\frac{1}{x+1}+\frac{1}{2(x+2)}$$$$S\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)})x^n$$$$=\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n+1}+\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n+2}but$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}=-ln\mid 1-x\mid$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n+1}=\sum _{n=2}^{+\mathrm{\infty }}\frac{x^{n-1}}{n}=\frac{1}{x}\left(\sum _{n=2}^{+\mathrm{\infty }}\frac{x^n}{n}\right)$$$$=\frac{1}{x}(-ln\mid 1-x\mid -x)$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n+2}=\sum _{n=3}^{+\mathrm{\infty }}\frac{x^{n-2}}{n}=\frac{1}{x^2}\sum _{n=3}^{+\mathrm{\infty }}\frac{x^n}{n}$$$$=\frac{1}{x^2}(\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}-x-\frac{x^2}{2})$$$$=-\frac{ln\mid 1-x\mid }{x^2}-\frac{1}{x}-\frac{1}{2}\Rightarrow$$$$S\left(x\right)=-\frac{1}{2}ln\mid 1-x\mid -\frac{1}{x}(-ln\mid 1-x\mid -x)$$$$-\frac{ln\mid 1-x\mid }{2x^2}-\frac{1}{2x}-\frac{1}{4}$$$$S\left(x\right)=(-\frac{1}{2}+\frac{1}{x}-\frac{1}{2x^2})ln\mid 1-x\mid +1-\frac{1}{2x}-\frac{1}{4}$$$$S\left(x\right)=(-\frac{1}{2}+\frac{1}{x}-\frac{1}{2x^2})ln\mid 1-x\mid -\frac{1}{2x}+\frac{3}{4}.$$$$2)\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{2^{n}n(n+1)(n+2)}=S(-\frac{1}{2})$$$$=(-\frac{1}{2}-2-2)ln\left(\frac{3}{2}\right)+1+\frac{3}{4}$$$$=(-\frac{1}{2}-4)ln\left(\frac{3}{2}\right)+\frac{7}{4}$$$$=\frac{7}{4}-\frac{9}{2}(ln\left(3\right)-ln\left(2\right)).$$

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