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Question-108415




Question Number 108415 by mathdave last updated on 16/Aug/20
Commented by bobhans last updated on 17/Aug/20
(1/(n(n+2)(n+4))) = (p/n)+(q/(n+2))+(r/(n+4))  1=p(n+2)(n+4)+qn(n+4)+rn(n+2)  n=0⇒1=8p→p=(1/8)  n=−2⇒1=−4q→q=−(1/4)  n=−4⇒1=8r→r=(1/8)  Σ_(n=1) ^∞  (1/(8n))−(1/(4(n+2)))+(1/(8(n+4))) = I+J+H  H=(1/8)Σ_(n=1) ^∞ (1/(n+4))=(1/8)((1/5)+(1/6)+Σ_(n=3) ^∞  (1/(n+4)))  (∗)(1/8)Σ_(n=3) ^∞  (1/(n+4))=(1/8)Σ_(n=6) ^∞  (1/(n+1))  (∗∗)(1/8)Σ_(n=1) ^∞ (1/n)=(1/8)(1+(1/2)+(1/3)+(1/4)+(1/5)+Σ_(n=6) ^∞  (1/n))
$$\frac{\mathrm{1}}{{n}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{4}\right)}\:=\:\frac{{p}}{{n}}+\frac{{q}}{{n}+\mathrm{2}}+\frac{{r}}{{n}+\mathrm{4}} \\ $$$$\mathrm{1}={p}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{4}\right)+{qn}\left({n}+\mathrm{4}\right)+{rn}\left({n}+\mathrm{2}\right) \\ $$$${n}=\mathrm{0}\Rightarrow\mathrm{1}=\mathrm{8}{p}\rightarrow{p}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${n}=−\mathrm{2}\Rightarrow\mathrm{1}=−\mathrm{4}{q}\rightarrow{q}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${n}=−\mathrm{4}\Rightarrow\mathrm{1}=\mathrm{8}{r}\rightarrow{r}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{8}{n}}−\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{2}\right)}+\frac{\mathrm{1}}{\mathrm{8}\left({n}+\mathrm{4}\right)}\:=\:{I}+{J}+{H} \\ $$$${H}=\frac{\mathrm{1}}{\mathrm{8}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}+\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}+\mathrm{4}}\right) \\ $$$$\left(\ast\right)\frac{\mathrm{1}}{\mathrm{8}}\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}+\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{8}}\underset{{n}=\mathrm{6}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$\left(\ast\ast\right)\frac{\mathrm{1}}{\mathrm{8}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+\underset{{n}=\mathrm{6}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}}\right) \\ $$$$ \\ $$

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