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Question Number 175230 by infinityaction last updated on 24/Aug/22

Answered by Ar Brandon last updated on 24/Aug/22

S=Σ_(n=3) ^∞ (1/( ^n C_3 ))=Σ_(n=0) ^∞ ((n!×3!)/((n+3)!)) =6Σ_(n=0) ^∞ (1/((n+3)(n+2)(n+1))) =6Σ_(n=0) ^∞ ((1/(2(n+1)))−(1/(n+2))+(1/(2(n+3)))) =6((1/2)(T+1)−T+(1/2)(T−(1/2))) T=Σ_(n=0) ^∞ (1/(n+2))⇒Σ_(n=0) ^∞ (1/(n+1))=T+1, Σ_(n=0) ^∞ (1/(n+3))=T−(1/2) ; =6((1/2)−(1/4))=6((1/4))=(3/2)★

$${S}=\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\overset{{n}} {\:}{C}_{\mathrm{3}} }=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}!×\mathrm{3}!}{\left({n}+\mathrm{3}\right)!} \\ $$$$\:\:\:=\mathrm{6}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{3}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)} \\ $$$$\:\:\:=\mathrm{6}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{n}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{3}\right)}\right) \\ $$$$\:\:\:=\mathrm{6}\left(\frac{\mathrm{1}}{\mathrm{2}}\left({T}+\mathrm{1}\right)−{T}+\frac{\mathrm{1}}{\mathrm{2}}\left({T}−\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$\:\:\:{T}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{2}}\Rightarrow\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}={T}+\mathrm{1}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{3}}={T}−\frac{\mathrm{1}}{\mathrm{2}}\:; \\ $$$$\:\:\:=\mathrm{6}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{6}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\mathrm{3}}{\mathrm{2}}\bigstar \\ $$

Commented by infinityaction last updated on 24/Aug/22

sir explain red term Σ_(n=0) ^∞ ((n!×3!)/((n+3)!))=6Σ_(n=0) ^∞ (((n+1)!−n!)/((n+3)!))

$${sir}\:{explain}\:{red}\:{term} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}!×\mathrm{3}!}{\left({n}+\mathrm{3}\right)!}=\mathrm{6}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({n}+\mathrm{1}\right)!−{n}!}{\left({n}+\mathrm{3}\right)!} \\ $$$$\:\:\: \\ $$

Commented by infinityaction last updated on 24/Aug/22

okey sir thank you

$${okey}\:{sir}\:{thank}\:{you} \\ $$

Commented by Ar Brandon last updated on 24/Aug/22

Sorry, mistake. Edited!

Commented by Tawa11 last updated on 25/Aug/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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