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Category: Relation and Functions

U-n-is-a-sequence-wich-verify-n-N-U-n-U-n-1-1-n-1-calculate-U-n-interms-of-n-2-is-the-sequence-U-n-convergent-

Question Number 65488 by mathmax by abdo last updated on 30/Jul/19 $${U}_{{n}} {is}\:{a}\:{sequence}\:{wich}\:{verify}\:\:\forall{n}\in{N}^{\bigstar} \\ $$$${U}_{{n}} \:+{U}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{{n}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\:{U}_{{n}} \:{interms}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right)\:{is}\:{the}\:{sequence}\:{U}_{{n}} {convergent}? \\ $$…

U-n-is-a-sequence-wich-verify-U-n-U-n-1-1-n-2-1-find-U-n-interms-of-n-2-calculate-lim-n-U-n-

Question Number 65489 by mathmax by abdo last updated on 30/Jul/19 $${U}_{{n}} \:{is}\:{a}\:{sequence}\:{wich}\:{verify}\:\:{U}_{{n}} \:+{U}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right)\:{find}\:{U}_{{n}} \:{interms}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{lim}_{{n}\rightarrow+\infty} \:{U}_{{n}} \\ $$ Commented…

1-calculate-0-dx-1-e-nx-with-n-integr-natural-and-n-1-2-conclude-the-value-of-k-1-1-k-1-k-

Question Number 65383 by mathmax by abdo last updated on 29/Jul/19 $$\left.\mathrm{1}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\mathrm{1}+{e}^{{nx}} }\:\:\:{with}\:{n}\:{integr}\:{natural}\:\:{and}\:{n}\geqslant\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{conclude}\:{the}\:{value}\:{of}\:\sum_{{k}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}} \\ $$ Commented by mathmax…

let-U-n-a-sequence-wich-verify-U-n-U-n-1-U-n-2-n-1-n-for-all-integr-n-calculate-interms-of-n-A-n-k-0-n-1-k-U-k-the-first-term-is-U-0-

Question Number 65297 by mathmax by abdo last updated on 28/Jul/19 $${let}\:\:\:{U}_{{n}} \:\:{a}\:{sequence}\:{wich}\:{verify}\:\:{U}_{{n}} \:+{U}_{{n}+\mathrm{1}} +{U}_{{n}+\mathrm{2}} \:={n}\left(−\mathrm{1}\right)^{{n}} \\ $$$${for}\:{all}\:{integr}\:{n}\:\:\:{calculate}\:{interms}\:{of}\:{n} \\ $$$${A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{k}} \:{U}_{{k}} \\…