Question Number 29511 by abdo imad last updated on 09/Feb/18 $${find}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\:\:\:\frac{\left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}\right)!}\:. \\ $$ Commented by prof Abdo imad last updated on 12/Feb/18 $${let}\:{use}\:{stirling}\:{formula}\:{we}\:{have}…
Question Number 29510 by abdo imad last updated on 09/Feb/18 $${let}\:\:\:{w}_{{n}} =\:\frac{\mathrm{1}}{{n}}\left(\:\mathrm{1}\:+{e}^{\frac{\mathrm{1}}{{n}}} \:+{e}^{\frac{\mathrm{2}}{{n}}} \:+\:…\:{e}^{\frac{{n}−\mathrm{1}}{{n}}} \right)\:{find}\:{lim}_{{n}\rightarrow+\infty} {w}_{{n}} . \\ $$ Commented by abdo imad last updated…
Question Number 29509 by abdo imad last updated on 09/Feb/18 $${find}\:{lim}_{{n}\rightarrow+\:\infty} \:\:\:\:\frac{{n}!}{\mathrm{2}^{{n}−\mathrm{1}} }\:. \\ $$ Commented by abdo imad last updated on 11/Feb/18 $${let}\:{use}\:{stirling}\:{formula}\:\:{n}!\:\sim\:{n}^{{n}} \:{e}^{−{n}}…
Question Number 160576 by mnjuly1970 last updated on 02/Dec/21 Commented by aleks041103 last updated on 02/Dec/21 $${What}\:{do}\:{the}\:{square}\:{brakets}\:{mean}? \\ $$ Commented by mnjuly1970 last updated on…
Question Number 29505 by abdo imad last updated on 09/Feb/18 $${let}\:{give}\:{u}_{{n}} =\:\:\sum_{{k}=\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} \:\mathrm{2}^{{k}} }\:\:{find}\:\:{lim}_{{n}\rightarrow\infty} {u}_{{n}} \:\:. \\ $$ Commented by abdo imad last…
Question Number 29500 by abdo imad last updated on 09/Feb/18 $${nature}\:{of}\:{the}\:{sequence}\:\:{u}_{{n}} =\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{kln}\left({k}\right)}\:. \\ $$ Commented by abdo imad last updated on 11/Feb/18…
Question Number 29499 by abdo imad last updated on 09/Feb/18 $${find}\:\:{lim}_{{n}\rightarrow+\infty} \left(\:\:\frac{\mathrm{1}}{\:\sqrt{{n}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{n}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}{n}}}\:+….+\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} }}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 29461 by prof Abdo imad last updated on 09/Feb/18 $${let}\:{give}\:\:{u}_{{n}} =\:\frac{\mathrm{1}}{\:\sqrt{{n}}}\left(\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+…+\frac{\mathrm{1}}{\:\sqrt{{n}}}\right)\: \\ $$$${find}\:{lim}_{{n}\rightarrow+\infty} {u}_{{n}\:} . \\ $$ Commented by prof Abdo imad last…
Question Number 29460 by prof Abdo imad last updated on 08/Feb/18 $${find}\:\:\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{{e}^{\sqrt{\mathrm{1}+{sinx}}} \:\:−{e}}{{tanx}}. \\ $$ Answered by Cheyboy last updated on 09/Feb/18 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\sqrt{\mathrm{1}+\mathrm{sin}\:{x}}}…
Question Number 29459 by prof Abdo imad last updated on 08/Feb/18 $${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\left(\frac{{sinx}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{1}−{cosx}}} \:. \\ $$ Commented by Cheyboy last updated on 09/Feb/18 $$\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{e}}} \\…