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Question Number 146307 by mnjuly1970 last updated on 12/Jul/21

$$$$$${\mathrm{S}}_n=\underset{k=1}{\sum ^n}\frac{1}{k(k+2)k+4)}$$$${lim}_{n\to \mathrm{\infty }}\left({\mathrm{S}}_n\right)=?$$

Answered by mathmax by abdo last updated on 12/Jul/21

$$\mathrm{let}\mathrm{decompose}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}(\mathrm{x}+2)(\mathrm{x}+4)}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{x}+2}+\frac{\mathrm{c}}{\mathrm{x}+4}$$$$\mathrm{a}=\frac{1}{8},\mathrm{b}=\frac{1}{(-2)\left(2\right)}=-\frac{1}{4},\mathrm{c}=\frac{1}{(-4)(-2)}=\frac{1}{8}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{8\mathrm{x}}-\frac{1}{4(\mathrm{x}+2)}+\frac{1}{8(\mathrm{x}+4)}$$$${\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{f}\left(\mathrm{k}\right)=\frac{1}{8}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}-\frac{1}{4}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+2}+\frac{1}{8}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+4}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}},\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+2}=\sum _{\mathrm{k}=3}^{\mathrm{n}+2}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}+2}-\frac{3}{2}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+4}=\sum _{\mathrm{k}=5}^{\mathrm{n}+4}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}+4}-\frac{3}{2}-\frac{1}{3}-\frac{1}{4}={\mathrm{H}}_{\mathrm{n}+4}-\frac{3}{2}-\frac{7}{12}$$$$={\mathrm{H}}_{\mathrm{n}+4}-\frac{25}{12}\Rightarrow$$$${\mathrm{S}}_{\mathrm{n}}=\frac{1}{8}{\mathrm{H}}_{\mathrm{n}}-\frac{1}{4}({\mathrm{H}}_{\mathrm{n}+2}-\frac{3}{2})+\frac{1}{8}({\mathrm{H}}_{\mathrm{n}+4}-\frac{25}{12})$$$$=\frac{1}{8}({\mathrm{H}}_{\mathrm{n}}+{\mathrm{H}}_{\mathrm{n}+4})-\frac{1}{4}{\mathrm{H}}_{\mathrm{n}+2}+\frac{3}{8}-\frac{25}{96}$$$${\mathrm{S}}_{\mathrm{n}}\sim \frac{1}{8}(\mathrm{logn}+\gamma +\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)+\log (\mathrm{n}+4)+\gamma +\mathrm{o}\left(\frac{1}{\mathrm{n}+4}\right))$$$$-\frac{1}{4}(\log (\mathrm{n}+2)+\gamma +\mathrm{o}\left(\frac{1}{\mathrm{n}+2}\right))+\frac{3}{8}-\frac{25}{96}$$$${\mathrm{S}}_{\mathrm{n}}\sim \log \left(\frac{{({\mathrm{n}}^2+4\mathrm{n})}^{\frac{1}{8}}}{{(\mathrm{n}+2)}^{\frac{1}{4}}}\right)+\frac{3}{8}-\frac{25}{96}\mathrm{we}\mathrm{have}$$$${\lim }_{\mathrm{n}\to \mathrm{\infty }}\frac{{({\mathrm{n}}^2+4\mathrm{n})}^{\frac{1}{8}}}{{(\mathrm{n}+2)}^{\frac{1}{4}}}=1\Rightarrow {\lim }_{\mathrm{n}⌣\mathrm{\infty }}{\mathrm{S}}_{\mathrm{n}}=\frac{3}{8}-\frac{25}{96}=\frac{36-25}{96}=\frac{11}{96}$$

Commented by mnjuly1970 last updated on 12/Jul/21

$$thxmrmax...$$

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