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Question Number 100304 by Algoritm last updated on 26/Jun/20

Commented by PRITHWISH SEN 2 last updated on 26/Jun/20

$${\mathrm{t}}_{\mathrm{n}}=\frac{\mathrm{n}}{(\mathrm{n}+1)!}=\frac{1}{\mathrm{n}!}-\frac{1}{(\mathrm{n}+1)!}$$$$\mathrm{and}\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{telescopic}\mathrm{series}$$

Commented by Dwaipayan Shikari last updated on 26/Jun/20

$${\mathrm{T}}_{\mathrm{n}}=\frac{\mathrm{n}}{(\mathrm{n}+1)!}=\frac{1}{\mathrm{n}!}-\frac{1}{(\mathrm{n}+1)!}$$$$\mathrm{\Sigma }{\mathrm{T}}_{\mathrm{n}}=\mathrm{\Sigma }\frac{1}{\mathrm{n}!}-\frac{1}{(\mathrm{n}+1)!}$$$${\mathrm{S}}_{\mathrm{n}}=1-\frac{1}{100!}\{\mathrm{General}\mathrm{sum}=1-\frac{1}{(\mathrm{n}+1)!}$$$$Supposeyoutake1$$$$thenitsumis1-\frac{1}{2!}=\frac{1}{2}\left\{whichis\frac{1}{2!}\right\}$$$$taking21-\frac{1}{3!}=\frac{5}{6}\{whichis\frac{1}{2!}+\frac{2}{3!}=\frac{5}{6}\}$$$$taking31-\frac{1}{4!}=\frac{23}{24}\{whichis\frac{5}{6}+\frac{3}{4!}=\frac{46}{48}=\frac{23}{24}$$

Commented by PRITHWISH SEN 2 last updated on 26/Jun/20

$$\mathrm{I}\mathrm{think}\mathrm{you}\mathrm{have}\mathrm{to}\mathrm{start}\mathrm{from}\mathrm{n}=2.\mathrm{Now}\mathrm{do}\mathrm{it}.$$

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