Question Number 138823 by mnjuly1970 last updated on 18/Apr/21
$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:……\mathscr{N}{ice}\:\:…\:\mathscr{C}{alculus}…… \\ $$$$\:\:\:\:\:\:\:\mathscr{E}{valuate}::\:\:\:\:\:\boldsymbol{\phi}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{}=? \\ $$
Answered by qaz last updated on 18/Apr/21
$$\phi=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\begin{pmatrix}{{n}+\mathrm{2}}\\{\mathrm{3}}\end{pmatrix}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{6}}{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){n}} \\ $$$$=\mathrm{6}\int_{\mathrm{0}} ^{\mathrm{1}} {dx}\int_{\mathrm{0}} ^{{x}} {dx}\int_{\mathrm{0}} ^{{x}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{{n}−\mathrm{1}} {dx} \\ $$$$=\mathrm{6}\int_{\mathrm{0}} ^{\mathrm{1}} {dx}\int_{\mathrm{0}} ^{{x}} {dx}\int_{\mathrm{0}} ^{{x}} \frac{{dx}}{\mathrm{1}−{x}} \\ $$$$=−\mathrm{6}\int_{\mathrm{0}} ^{\mathrm{1}} {dx}\int_{\mathrm{0}} ^{{x}} {ln}\left(\mathrm{1}−{x}\right){dx} \\ $$$$=\mathrm{6}\int_{\mathrm{0}} ^{\mathrm{1}} {xlnx}+\mathrm{1}−{xdx} \\ $$$$=\frac{\mathrm{9}}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 18/Apr/21
$${grateful}… \\ $$