Question Number 676 by 123456 last updated on 22/Feb/15 $${given}\:{two}\:{sequence}\:{a}_{{n}} >\mathrm{0},{b}_{{n}} >\mathrm{0}\:\:{such} \\ $$$$\forall{n}\in\mathbb{N}^{\ast} ,{a}_{{n}} ^{{n}} <{b}_{{n}} <{a}_{{n}} ^{\mathrm{1}/{n}} \\ $$$${a}.\:{if}\:\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}{a}_{{n}} \:{converge}\:{then}\:\underset{{n}=\mathrm{1}} {\overset{+\infty}…
Question Number 585 by 123456 last updated on 02/Feb/15 $${f}_{{n}} \left({x}\right)=\left(\mathrm{1}+\frac{{x}}{{n}}\right)^{{nx}} ,{n}\in\mathbb{N}^{\ast} ,{x}\in\mathbb{R},{x}\geqslant\mathrm{0} \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{f}_{{n}} \left({x}\right)= \\ $$ Answered by prakash jain last updated…
Question Number 561 by 123456 last updated on 27/Jan/15 $${a}_{{n}+\mathrm{2}} =\frac{{a}_{{n}+\mathrm{1}} +{a}_{{n}} +{a}_{{n}+\mathrm{1}} {a}_{{n}} }{\mathrm{3}},{a}_{\mathrm{0}} ={e},{a}_{\mathrm{1}} =\pi \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} =? \\ $$ Terms of…
Question Number 66084 by F_Nongue last updated on 09/Aug/19 $${How}\:{to}\:{solve}\:{this}\:{limit}? \\ $$$$\underset{{x}\rightarrow\infty} {{lim}}\left(\mathrm{7}{x}+\frac{\mathrm{2}}{{x}}\right)^{{x}} \\ $$ Commented by mathmax by abdo last updated on 09/Aug/19 $${let}\:{A}\left({x}\right)\:=\left(\mathrm{7}{x}+\frac{\mathrm{2}}{{x}}\right)^{{x}}…
Question Number 494 by 123456 last updated on 15/Jan/15 $${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…