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Question Number 32042 Answers: 0 Comments: 1
$${let}\:{u}_{{n}} \:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}^{{n}} \:{sin}\left(\pi{x}\right){dx} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\Sigma\:{u}_{{n}} \:{converges} \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:\Sigma\:{u}_{{n}} =\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{sint}}{{t}}{dt}\:. \\ $$
Question Number 32041 Answers: 0 Comments: 1
$${find}\:{the}\:{nature}\:{of}\:\Sigma\:{u}_{{n}} \:\:/ \\ $$$${u}_{{n}} =\:\frac{\sqrt{\mathrm{1}}\:+\sqrt{\mathrm{2}}\:+....+\sqrt{{n}}}{{n}^{\mathrm{3}} }\:. \\ $$
Question Number 32040 Answers: 0 Comments: 2
$${let}\:{give}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{dt}}{\mathrm{1}+{x}\:{tant}}\:\: \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{tant}}{\left(\mathrm{1}+{xtant}\right)^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right){give}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{tant}}{\left(\mathrm{1}+\sqrt{\mathrm{3}}\:{tant}\right)^{\mathrm{2}} }\:{dt}\:. \\ $$
Question Number 32039 Answers: 0 Comments: 3
$${a}>−\mathrm{1}\:\:{calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{dt}}{\mathrm{1}+{a}\:{tan}^{\mathrm{2}} {t}}\:. \\ $$$$\left.\mathrm{2}\right)\:{find}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{tan}^{\mathrm{2}} {t}}{\left(\mathrm{1}+{atan}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }\:{dt} \\ $$$$\left.\mathrm{3}\right)\:\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{tan}^{\mathrm{2}} {t}}{\left(\mathrm{1}+\mathrm{2}{tan}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }{dt}.\: \\ $$
Question Number 32037 Answers: 0 Comments: 1
$${let}\:\:{u}_{{n}} ={cos}\left(\pi\sqrt{{n}^{\mathrm{2}} \:+{n}+\mathrm{1}}\right)\:{find}\:{nature}\:{of}\:\Sigma\:{u}_{{n}} . \\ $$$$ \\ $$
Question Number 32036 Answers: 0 Comments: 0
$${nature}\:{of}\:\Sigma\:{u}_{{n}} \:\:{with}\:{u}_{{n}} =\:\:\:\frac{\mathrm{1}}{\left({ln}\left(\mathrm{2}\right)\right)^{\mathrm{2}} \:+....+\left({ln}\left({n}\right)\right)^{\mathrm{2}} }\:\:. \\ $$
Question Number 32034 Answers: 0 Comments: 0
$${let}\:{u}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\mathrm{1}+{x}+...+{x}^{{n}} }\:\:{study}\:{the}\:{convergence}\:{of} \\ $$$$\Sigma\:{u}_{{n}} \:\:. \\ $$
Question Number 32033 Answers: 0 Comments: 0
$${let}\:{consider}\:{the}\:{sequence}\:\:\left({u}_{{n}} \right)\:\:/{u}_{\mathrm{0}} \in\left[\mathrm{0},\mathrm{1}\right]\:{and} \\ $$$$\forall{n}\in{N}\:\:{u}_{{n}+\mathrm{1}} =\:{u}_{{n}} \:−{u}_{{n}} ^{\mathrm{2}} \\ $$$$\left.\mathrm{1}\right)\:{give}\:{a}\:{simple}\:{equivalent}\:{of}\:\:{u}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{nature}\:{of}\:\Sigma\:{u}_{{n}} . \\ $$
Question Number 32031 Answers: 0 Comments: 1
$${let}\:\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{ax}} {ln}\left({x}\right){dx}\:\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{f}\left({a}\right)\: \\ $$$$\left.\mathrm{2}\right)\:\:{find}\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{ax}} \left({xlnx}\right){dx} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\mathrm{2}{x}} \left({xlnx}\right){dx}\:\:. \\ $$
Question Number 32029 Answers: 0 Comments: 0
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\alpha{x}} {ln}\left({x}\right)\:{dx}\:\:{with}\:\:\alpha>\mathrm{0}\:. \\ $$
Question Number 32028 Answers: 1 Comments: 0
$${If}\:\frac{\mathrm{2}{z}_{\mathrm{1}} }{\mathrm{3}{z}_{\mathrm{2}} }\:{is}\:{a}\:{purely}\:{imaginary}\:{number}, \\ $$$${then}\:{find}\:{the}\:{value}\:{of}\:\mid\frac{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} }{{z}_{\mathrm{1}} +{z}_{\mathrm{2}} }\mid\:. \\ $$
Question Number 32026 Answers: 0 Comments: 1
$${let}\:\alpha>\mathrm{0}\:{prove}\:{that}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\alpha}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{\alpha−\mathrm{1}} }{\mathrm{1}+{x}}{dx}\:. \\ $$
Question Number 32025 Answers: 0 Comments: 2
$${calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{1}}\:. \\ $$
Question Number 32008 Answers: 1 Comments: 3
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left[\left(\frac{\mathrm{1}}{{n}}\right)^{{n}} +\left(\frac{\mathrm{2}}{{n}}\right)^{{n}} +..+\left(\frac{{n}}{{n}}\right)^{{n}} \right]=... \\ $$
Question Number 32002 Answers: 1 Comments: 0
$${If}\:\:\boldsymbol{{z}}^{\mathrm{3}} =\bar {\boldsymbol{{z}}}\:{prove}\: \\ $$$${then}\:\mid\boldsymbol{{z}}\mid=\mathrm{1}. \\ $$
Question Number 31991 Answers: 0 Comments: 1
$${g}_{{n}} =\sqrt{{g}_{{n}−\mathrm{1}} +{g}_{{n}−\mathrm{2}} } \\ $$$${g}_{\mathrm{1}} =\mathrm{1} \\ $$$${g}_{\mathrm{2}} =\mathrm{3} \\ $$$${g}_{{n}} =.. \\ $$
Question Number 31990 Answers: 1 Comments: 3
$${a}_{{n}} =\mathrm{2}{a}_{{n}−\mathrm{1}} +\mathrm{3}{a}_{{n}−\mathrm{2}} \\ $$$${a}_{\mathrm{0}} =\mathrm{1} \\ $$$${a}_{\mathrm{1}} =\mathrm{2} \\ $$$${a}_{{n}} =... \\ $$
Question Number 31984 Answers: 0 Comments: 1
$${study}\:{the}\:{covergence}\:{of}\:\:\Sigma\:{u}_{{n}} \:\:{with} \\ $$$${u}_{{n}} =^{{n}} \sqrt{\frac{{n}}{{n}+\mathrm{1}}}\:−\mathrm{1}\:\:\:. \\ $$
Question Number 31983 Answers: 0 Comments: 2
$${calculate}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{n}^{\mathrm{2}} \:−\mathrm{2}}{{n}!}\:\:. \\ $$
Question Number 31982 Answers: 0 Comments: 2
$${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}\:. \\ $$
Question Number 31981 Answers: 0 Comments: 1
$${find}\:{the}\:{nature}\:{of}\:\:\:\sum_{{n}\geqslant\mathrm{2}} \:\frac{\mathrm{1}}{{nln}\left({n}\right)}\:. \\ $$
Question Number 31980 Answers: 0 Comments: 0
$${let}\:−\mathrm{1}<{x}<\mathrm{1}\:{calculate}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{{x}^{{n}} }{\left(\mathrm{1}−{x}^{{n}} \right)\left(\mathrm{1}−{x}^{{n}+\mathrm{1}} \right)}\:\:. \\ $$
Question Number 31979 Answers: 0 Comments: 1
$${calculate}\:\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:\:. \\ $$
Question Number 31978 Answers: 1 Comments: 0
$${find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\mathrm{2}{n}+\mathrm{5}\right)}. \\ $$
Question Number 31977 Answers: 1 Comments: 2
$${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$
Question Number 31976 Answers: 0 Comments: 0
$${let}\:{u}_{{n}} =^{{n}+\mathrm{1}} \sqrt{{n}+\mathrm{1}}\:−\:^{{n}} \sqrt{{n}}\:\:{find}\:{radius}\:{of}\:{convergence}\: \\ $$$${for}\:\:\Sigma\:{u}_{{n}} {z}^{{n}} \:\:\:\:\left({z}\in{C}\right). \\ $$
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