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Question Number 99118 by pticantor last updated on 18/Jun/20

Answered by mathmax by abdo last updated on 18/Jun/20

$$1)\mathrm{let}\mathrm{p}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{k}}\Rightarrow {\mathrm{p}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{kx}}^{\mathrm{k}-1}\Rightarrow \sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{k}{\mathrm{x}}^{\mathrm{k}-1}=$$$$=\frac{\mathrm{d}}{\mathrm{dx}}(\frac{{\mathrm{x}}^{\mathrm{n}}-1}{\mathrm{x}-1}-1)=\frac{{\mathrm{nx}}^{\mathrm{n}+1}-(\mathrm{n}+1){\mathrm{x}}^{\mathrm{n}}+1}{{(\mathrm{x}-1)}^2}\Rightarrow$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{k}{2}^{\mathrm{k}-1}=\mathrm{n}{2}^{\mathrm{n}+1}-(\mathrm{n}+1)2^{\mathrm{n}}+1=(2\mathrm{n}-\mathrm{n}-1)2^{\mathrm{n}}+1=(\mathrm{n}-1)2^{\mathrm{n}}+1$$

Answered by mathmax by abdo last updated on 18/Jun/20

$$\mathrm{b})\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{k}(\mathrm{n}-\mathrm{k})=\mathrm{n}\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{k}-\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{k}}^2$$$$=\mathrm{n}\frac{\mathrm{n}(\mathrm{n}+1)}{2}-\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\{\mathrm{n}-\frac{2\mathrm{n}+1}{3}\}$$$$=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left(\frac{\mathrm{n}-1}{3}\right)=\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}-1)}{6}$$

Answered by mathmax by abdo last updated on 19/Jun/20

$$2)\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{\mathrm{n}(\mathrm{n}+1)}={\lim }_{\mathrm{n}\to +\mathrm{\infty }}\sum _{\mathrm{k}=1}^{\mathrm{n}}(\frac{1}{\mathrm{k}}-\frac{1}{\mathrm{k}+1})={\lim }_{\mathrm{n}\to +\mathrm{\infty }}(1-\frac{1}{\mathrm{n}+1})=1$$

Answered by mathmax by abdo last updated on 19/Jun/20

$$2^{\mathrm{b})}\mathrm{let}{\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)}\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}(\mathrm{x}+1)(\mathrm{x}+2)}$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{x}+1}+\frac{\mathrm{c}}{\mathrm{x}+2}$$$$\mathrm{a}=\frac{1}{2},\mathrm{b}=-1,\mathrm{c}=\frac{1}{2}\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{1}{2\mathrm{x}}-\frac{1}{\mathrm{x}+1}+\frac{1}{2(\mathrm{x}+2)}\Rightarrow$$$${\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{2\mathrm{k}}-\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+1}+\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{2(\mathrm{k}+2)}\mathrm{we}\mathrm{have}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}},\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+1}=\sum _{\mathrm{k}=2}^{\mathrm{n}+1}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}+1}-1$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+2}=\sum _{\mathrm{k}=3}^{\mathrm{n}+2}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}+2}-\frac{3}{2}\Rightarrow {\mathrm{S}}_{\mathrm{n}}=\frac{1}{2}{\mathrm{H}}_{\mathrm{n}}-{\mathrm{H}}_{\mathrm{n}+1}+1+\frac{1}{2}{\mathrm{H}}_{\mathrm{n}+2}-\frac{3}{4}$$$$=\frac{1}{2}({\mathrm{H}}_{\mathrm{n}}+{\mathrm{H}}_{\mathrm{n}+2})-{\mathrm{H}}_{\mathrm{n}+1}+\frac{1}{4}=\frac{1}{2}(\ln \left(\mathrm{n}\right)+\gamma +\ln (\mathrm{n}+2)+\gamma +\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)$$$$-\ln (\mathrm{n}+1)-\gamma -\mathrm{o}\left(\frac{1}{\mathrm{n}+1}\right)+\frac{1}{4}=\ln \left(\sqrt{{\mathrm{n}}^2+2\mathrm{n}}\right)-\ln (\mathrm{n}+1)+\frac{1}{4}+\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)$$$$=\ln \left(\frac{\sqrt{{\mathrm{n}}^2+2\mathrm{n}}}{\mathrm{n}+1}\right)+\frac{1}{4}+\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)\to \frac{1}{4}\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{S}}_{\mathrm{n}}=\frac{1}{4}$$

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