Question Number 31533 by abdo imad last updated on 09/Mar/18 $${let}\:{put}\:{S}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+{x}} \\ $$$$\left.\mathrm{1}\left.\right)\:{prove}\:{that}\:{S}\:{is}\:{C}^{\mathrm{1}} \:{on}\right]\mathrm{0}'+\infty\left[\right. \\ $$$$\left.\mathrm{2}\right){give}\:{the}\:{variation}\:{of}\:{S}\left({x}\right) \\ $$$$\left.\mathrm{3}\right){prove}\:{that}\:\forall{x}>\mathrm{0}\:{S}\left({x}+\mathrm{1}\right)+{S}\left({x}\right)=\frac{\mathrm{1}}{{x}} \\ $$$$\left.\mathrm{4}\right){give}\:{a}\:{equivalent}\:{for}\:{S}\:{at}\:\mathrm{0} \\ $$$$\left.\mathrm{5}\right){find}\:{a}\:{equivalent}\:{for}\:{S}\:{at}\:+\infty.…
Question Number 31526 by abdo imad last updated on 09/Mar/18 $$\left.\mathrm{1}\right){find}\:{lim}_{{n}\rightarrow\infty} \left(\:\frac{{a}^{\frac{\mathrm{1}}{{n}}} \:+{b}^{\frac{\mathrm{1}}{{n}}} }{\mathrm{2}}\right)^{{n}} \\ $$$$\left.\mathrm{2}\right)\:{let}\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}}\:{calculate}\:{lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left(\:^{{n}} \sqrt{{cos}\theta}\:+^{{n}} \sqrt{{sin}\theta}\:\right)^{{n}} \\ $$ Commented by abdo…
Question Number 31523 by abdo imad last updated on 09/Mar/18 $${let}\:{give}\:{u}_{{n}} =^{{n}+\mathrm{1}} \sqrt{{n}+\mathrm{1}}\:\:−^{{n}} \sqrt{{n}}\: \\ $$$$\left.\mathrm{1}\right)\:{study}\:{the}\:{convergence}\:{of}\:\left({u}_{{n}} \right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{nature}\:{of}\:{serie}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} {u}_{{n}} \\ $$ Terms of…
Question Number 31521 by abdo imad last updated on 09/Mar/18 $${study}\:{the}\:{convergence}\:{of}\:{u}_{{n}} =\sqrt{{n}+\mathrm{1}}\:−\sqrt{{n}−\mathrm{1}}\: \\ $$ Commented by abdo imad last updated on 12/Mar/18 $${we}\:{have}\:{u}_{{n}} =\sqrt{{n}}\left(\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}}}\:−\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{n}}}\:\right)\:{but} \\…
Question Number 31509 by abdo imad last updated on 09/Mar/18 $${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} \:+\mathrm{2}{kn}}}\:\:{find}\:\:{lim}_{{n}\rightarrow\infty} \:{S}_{{n}} . \\ $$ Commented by abdo imad last updated…
Question Number 31510 by abdo imad last updated on 09/Mar/18 $$\left.{find}\:{lim}_{{n}\rightarrow\infty} \:\:\:^{{n}} \sqrt{\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right.}\right) \\ $$ Commented by abdo imad last updated on…
Question Number 31508 by abdo imad last updated on 09/Mar/18 $${let}\:{give}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\sqrt{{k}}\:\:{find}\:{a}\:{simple}\:{eqivalent}\:{of}\:{S}_{{n}} . \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 31500 by abdo imad last updated on 09/Mar/18 $${let}\:{L}_{{n}} \left({x}\right)=\:{e}^{{x}} \:\left({e}^{−{x}} \:{x}^{{n}} \right)^{\left({n}\right)} \: \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{L}_{{n}} \:{is}\:{a}\:{polynomial} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{degL}_{{n}\:} {and}\:{the}\:{leading}\:{coefficient}\:. \\ $$ Commented…
Question Number 162509 by mathmax by abdo last updated on 30/Dec/21 $$\mathrm{find}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{3}} \left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 31422 by abdo imad last updated on 08/Mar/18 $${find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{arctan}\left(\frac{\mathrm{2}}{\mathrm{16}{n}^{\mathrm{2}} \:+\mathrm{8}{n}−\mathrm{2}}\right). \\ $$ Commented by rahul 19 last updated on 08/Mar/18 $${can}\:{u}\:{plz}\:{provide}\:{sol}.\:{for}\:{this}\:{one}\:.…