Question Number 494 by 123456 last updated on 15/Jan/15 $${a}_{\mathrm{0}} =\mathrm{0} \\ $$$${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{{n}} =\frac{{a}_{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} +{a}_{{n}−\mathrm{1}} {a}_{{n}−\mathrm{2}} }{\mathrm{3}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} =…
Question Number 131548 by liberty last updated on 06/Feb/21 $$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{n}}\:\underset{\mathrm{0}} {\overset{\:\:\frac{\pi}{\mathrm{2}}} {\int}}\:\mathrm{cos}\:^{\mathrm{2n}+\mathrm{1}} \left(\theta\right)\:\mathrm{d}\theta\:=? \\ $$ Answered by rs4089 last updated on 06/Feb/21 $$\frac{\pi\sqrt{\pi}}{\mathrm{2}} \\…
Question Number 66010 by jimful last updated on 07/Aug/19 $${Using}\:\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}!}={e}\:,\:{prove}\:{that} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} ={e} \\ $$ Commented by Prithwish sen last updated on…