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Question Number 138823 by mnjuly1970 last updated on 18/Apr/21

$$$$$$......\mathcal{N}ice...\mathcal{C}alculus......$$$$\mathcal{E}valuate::\mathit{\varphi }=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{}=?$$

Answered by qaz last updated on 18/Apr/21

$$\varphi =\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\left(\begin{array}{c}n+2\\ 3\end{array}\right)}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{6}{(n+2)(n+1)n}$$$$=6{\int }_0^{1}dx{\int }_0^{x}dx{\int }_0^x\underset{n=1}{\sum ^{\mathrm{\infty }}}{x}^{n-1}dx$$$$=6{\int }_0^{1}dx{\int }_0^{x}dx{\int }_0^x\frac{dx}{1-x}$$$$=-6{\int }_0^{1}dx{\int }_0^{x}ln(1-x)dx$$$$=6{\int }_0^{1}xlnx+1-xdx$$$$=\frac{9}{2}$$

Commented by mnjuly1970 last updated on 18/Apr/21

$$grateful...$$

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