Question Number 107287 by mathmax by abdo last updated on 09/Aug/20 $$\mathrm{find}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \sum_{\mathrm{k}=\mathrm{n}} ^{\mathrm{2n}−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{k}+\mathrm{n}} \\ $$ Answered by mathmax by abdo last updated on…
Question Number 107284 by mathmax by abdo last updated on 09/Aug/20 $$\mathrm{find}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\:\sqrt{\left(\mathrm{k}+\mathrm{n}\right)\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 41705 by abdo.msup.com last updated on 11/Aug/18 $${find}\:{nsture}\:{of}\:{the}\:{serie}\:\Sigma\:{u}_{{n}} \\ $$$${with}\:{u}_{{n}} =^{{n}} \sqrt{\frac{{n}}{{n}+\mathrm{1}}}\:\:\:−\mathrm{1} \\ $$ Commented by maxmathsup by imad last updated on 12/Aug/18…
Question Number 41706 by abdo.msup.com last updated on 11/Aug/18 $${study}\:{the}\:{convergence}\:{of}\: \\ $$$${u}_{{n}} =\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{\mathrm{1}}{{kln}\left({k}\right)}\:−{ln}\left({ln}\left({n}\right)\right. \\ $$ Commented by math khazana by abdo last updated…
Question Number 41704 by abdo.msup.com last updated on 11/Aug/18 $${let}\:{u}_{{n}} =\:\frac{\mathrm{1}}{\:\sqrt{{n}−\mathrm{1}}}\:+\frac{\mathrm{2}}{\:\sqrt{{n}}}\:+\frac{\mathrm{1}}{\:\sqrt{{n}+\mathrm{1}}} \\ $$$${calculate}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:{u}_{{n}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 41672 by math khazana by abdo last updated on 11/Aug/18 $${find}\:\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{3}{k}+\mathrm{1}}\:{interms}\:{of}\:{H}_{{n}} \\ $$ Commented by math khazana by abdo last updated…
Question Number 107147 by mathmax by abdo last updated on 09/Aug/20 $$\mathrm{find}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}^{+} } \:\:\:\:\frac{\int_{\mathrm{0}} ^{\mathrm{x}} \:\mathrm{e}^{\mathrm{t}} \mathrm{ln}\left(\mathrm{t}\right)\mathrm{dt}}{\mathrm{e}^{\mathrm{x}} \mathrm{lnx}} \\ $$ Terms of Service Privacy Policy…
Question Number 107144 by mathmax by abdo last updated on 09/Aug/20 $$\mathrm{solve}\:\mathrm{x}^{\mathrm{2}} \mathrm{y}^{''} −\mathrm{xy}^{'} \:+\mathrm{2y}\:=\mathrm{xe}^{−\mathrm{x}} \mathrm{sin}\left(\mathrm{2x}\right) \\ $$ Answered by bemath last updated on 09/Aug/20…
Question Number 41517 by maxmathsup by imad last updated on 08/Aug/18 $${let}\:\:{S}_{{n}} =\:\mathrm{1}\:+\frac{\mathrm{1}}{\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\right)}\:+\:\frac{\mathrm{1}}{\left(^{\mathrm{3}} \sqrt{\mathrm{3}}\right)}\:+\:….+\frac{\mathrm{1}}{\left(^{\mathrm{3}} \sqrt{{n}}\right)} \\ $$$${calculate}\:{lim}\:_{{n}\rightarrow+\infty} \:{S}_{{n}} \\ $$ Commented by maxmathsup by…
Question Number 41512 by maxmathsup by imad last updated on 08/Aug/18 $${let}\:\:{u}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }\:+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\:+….+\frac{\mathrm{1}}{{n}^{\mathrm{3}} } \\ $$$$\left.\mathrm{1}\right){prove}\:{that}\:\:\:\frac{\mathrm{9}}{\mathrm{8}}\:−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\leqslant\:{u}_{{n}} \leqslant\:\frac{\mathrm{3}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:\forall\:{n}\in{N}^{\bigstar} \:\:\:\mathrm{1}\leqslant{u}_{{n}} \leqslant\:\frac{\mathrm{3}}{\mathrm{2}}…