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Question Number 147572 by liberty last updated on 22/Jul/21

$$\sum _{n⩾1}\frac{4n-3}{(n^2+2n)(n+3)}=?$$

Answered by mathmax by abdo last updated on 22/Jul/21

$$\mathrm{let}\mathrm{S}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{4\mathrm{n}-3}{\mathrm{n}(\mathrm{n}+2)(\mathrm{n}+3)}\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{4\mathrm{x}-3}{\mathrm{x}(\mathrm{x}+2)(\mathrm{x}+3)}$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{x}+2}+\frac{\mathrm{c}}{\mathrm{x}+3}$$$$\mathrm{a}=\frac{-3}{6}=-\frac{1}{2},\mathrm{b}=\frac{-11}{(-2)\left(1\right)}=\frac{11}{2},\mathrm{c}=\frac{-15}{(-3)(-1)}=-5\Rightarrow$$$$\mathrm{F}\left(\mathrm{x}\right)=-\frac{1}{2\mathrm{x}}+\frac{11}{2(\mathrm{x}+2)}-\frac{5}{\mathrm{x}+3}\Rightarrow$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{F}\left(\mathrm{k}\right)=-\frac{1}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}+\frac{11}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+2}-5\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+3}={\mathrm{S}}_{\mathrm{n}}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}},\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+2}=\sum _{\mathrm{k}=3}^{\mathrm{n}+2}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}+2}-\frac{3}{2}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+3}=\sum _{\mathrm{k}=1}^{\mathrm{n}+4}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}+4}-\frac{3}{2}-\frac{1}{3}={\mathrm{H}}_{\mathrm{n}+4}-\frac{11}{6}\Rightarrow$$$${\mathrm{S}}_{\mathrm{n}}=-\frac{1}{2}{\mathrm{H}}_{\mathrm{n}}+\frac{11}{2}({\mathrm{H}}_{\mathrm{n}+2}-\frac{3}{2})-5({\mathrm{H}}_{\mathrm{n}+4}-\frac{11}{6})$$$$=-\frac{1}{2}{\mathrm{H}}_{\mathrm{n}}+\frac{11}{2}{\mathrm{H}}_{\mathrm{n}+2}-\frac{33}{4}+{5\mathrm{H}}_{\mathrm{n}+4}-\frac{55}{6}$$$${\mathrm{S}}_{\mathrm{n}}=-\frac{1}{2}(\mathrm{logn}+\gamma +\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)+\frac{11}{2}(\log (\mathrm{n}+2)+\gamma +\mathrm{o}\left(\frac{1}{\mathrm{n}+2}\right))$$$$-5(\log (\mathrm{n}+4)+\gamma +\mathrm{o}\left(\frac{1}{\mathrm{n}+4}\right))-\frac{33}{4}+\frac{55}{6}\Rightarrow$$$${\mathrm{S}}_{\mathrm{n}}\sim -\frac{1}{2}\mathrm{logn}-\frac{\gamma }{2}+\frac{11}{2}\mathrm{logn}+\frac{11}{2}\log (1+\frac{2}{\mathrm{n}})+\frac{11\gamma }{2}-5\log \left(\mathrm{n}\right)-5\log (1+\frac{4}{\mathrm{n}})$$$$-5\gamma -\frac{33}{4}+\frac{55}{6}\Rightarrow {\mathrm{S}}_{\mathrm{n}}\sim \frac{11}{2}\log (1+\frac{2}{\mathrm{n}})-5\log (1+\frac{4}{\mathrm{n}})-\frac{33}{4}+\frac{55}{6}\Rightarrow$$$${\lim }_{\mathrm{n}\to \mathrm{\infty }}{\mathrm{S}}_{\mathrm{n}}=\frac{55}{6}-\frac{33}{4}$$

Answered by qaz last updated on 22/Jul/21

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{4\mathrm{n}-3}{\mathrm{n}(\mathrm{n}+2)(\mathrm{n}+3)}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}(-\frac{1}{2\mathrm{n}}+\frac{11}{2}\cdot \frac{1}{\mathrm{n}+2}-5\cdot \frac{1}{\mathrm{n}+3})$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}(-\frac{1}{2}\cdot \frac{1}{\mathrm{n}+1}+\frac{11}{2}\cdot \frac{1}{\mathrm{n}+3}-5\cdot \frac{1}{\mathrm{n}+4})$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}(-\frac{1}{2}{\int }_0^1{\mathrm{x}}^{\mathrm{n}}\mathrm{dx}+\frac{11}{2}{\int }_0^1{\mathrm{x}}^{\mathrm{n}+2}\mathrm{dx}-5{\int }_0^1{\mathrm{x}}^{\mathrm{n}+3}\mathrm{dx})$$$$=-\frac{1}{2}{\int }_0^1\frac{\mathrm{dx}}{1-\mathrm{x}}\mathrm{dx}+\frac{11}{2}{\int }_0^1\frac{{\mathrm{x}}^2}{1-\mathrm{x}}\mathrm{dx}-5{\int }_0^1\frac{{\mathrm{x}}^3}{1-\mathrm{x}}\mathrm{dx}$$$$=\frac{1}{2}{\int }_0^1\frac{-1+{11\mathrm{x}}^2-{10\mathrm{x}}^3}{1-\mathrm{x}}\mathrm{dx}$$$$=-\frac{1}{2}{\int }_0^1(1+\mathrm{x}-{10\mathrm{x}}^2)\mathrm{dx}$$$$=\frac{11}{12}$$

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