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Question-203771

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Question Number 203771 by Calculusboy last updated on 27/Jan/24
Answered by DwaipayanShikari last updated on 27/Jan/24
Σ_(n=0) ^∞ (1/ (((n+3)),(3) )) =Σ_(n=0) ^∞ ((n!)/((n+3)!3!)) =(1/(3!))Σ_(n=0) ^∞ (1/((n+1)(n+2)(n+3))) =(1/(3!))Σ((1/(n+1))−(1/(n+2)))−(1/(2.3!))Σ((1/(n+1))−(1/(n+3))) =(1/(3!))(1−(1/2)+..)−(1/(2.3!))×1 =(1/(2.3!))=(1/(12))
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\begin{pmatrix}{{n}+\mathrm{3}}\\{\mathrm{3}}\end{pmatrix}} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}!}{\left({n}+\mathrm{3}\right)!\mathrm{3}!} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}!}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}!}\Sigma\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}.\mathrm{3}!}\Sigma\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{3}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}!}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+..\right)−\frac{\mathrm{1}}{\mathrm{2}.\mathrm{3}!}×\mathrm{1} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}.\mathrm{3}!}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$
Commented by aleks041103 last updated on 31/Jan/24
Σ_(n=0) ^∞ ((1/(n+1))−(1/(n+2)))=Σ_(n=1) ^∞ (1/n)−Σ_(n=2) ^∞ (1/n)=(1/1)=1 Σ_(n=0) ^∞ ((1/(n+1))−(1/(n+3)))=Σ_(n=1) ^∞ (1/n)−Σ_(n=3) ^∞ (1/n)= =(1/1)+(1/2)=(3/2) ⇒Σ_(n=0) ^∞ (1/ (((n+3)),(3) ))=(1/(3!))(Σ((1/(n+1))−(1/(n+2)))−(1/2)Σ((1/(n+1))−(1/(n+3))))= =(1/6)(1−(1/2) (3/2))=(1/6) (1/4) = (1/(24))
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}=\frac{\mathrm{1}}{\mathrm{1}}=\mathrm{1} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{3}}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\begin{pmatrix}{{n}+\mathrm{3}}\\{\mathrm{3}}\end{pmatrix}}=\frac{\mathrm{1}}{\mathrm{3}!}\left(\Sigma\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\Sigma\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{3}}\right)\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{6}}\:\frac{\mathrm{1}}{\mathrm{4}}\:=\:\frac{\mathrm{1}}{\mathrm{24}} \\ $$

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