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Question Number 203771 by Calculusboy last updated on 27/Jan/24

Answered by DwaipayanShikari last updated on 27/Jan/24

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\left(\begin{array}{c}n+3\\ 3\end{array}\right)}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{n!}{(n+3)!3!}$$$$=\frac{1}{3!}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(n+1)(n+2)(n+3)}$$$$=\frac{1}{3!}\mathrm{\Sigma }(\frac{1}{n+1}-\frac{1}{n+2})-\frac{1}{2.3!}\mathrm{\Sigma }(\frac{1}{n+1}-\frac{1}{n+3})$$$$=\frac{1}{3!}(1-\frac{1}{2}+..)-\frac{1}{2.3!}\times 1$$$$=\frac{1}{2.3!}=\frac{1}{12}$$

Commented by aleks041103 last updated on 31/Jan/24

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{n+1}-\frac{1}{n+2})=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{n}=\frac{1}{1}=1$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{n+1}-\frac{1}{n+3})=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\underset{n=3}{\sum ^{\mathrm{\infty }}}\frac{1}{n}=$$$$=\frac{1}{1}+\frac{1}{2}=\frac{3}{2}$$$$\Rightarrow \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\left(\begin{array}{c}n+3\\ 3\end{array}\right)}=\frac{1}{3!}(\mathrm{\Sigma }(\frac{1}{n+1}-\frac{1}{n+2})-\frac{1}{2}\mathrm{\Sigma }(\frac{1}{n+1}-\frac{1}{n+3}))=$$$$=\frac{1}{6}(1-\frac{1}{2}\frac{3}{2})=\frac{1}{6}\frac{1}{4}=\frac{1}{24}$$

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