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Category: Limits

x-i-1-i-n-n-real-number-positifs-wish-verfy-i-1-i-n-x-i-1-prove-that-1-i-n-x-i-2-1-n-

Question Number 26244 by abdo imad last updated on 22/Dec/17 $$\:\left({x}_{{i}} \:\right)_{\mathrm{1}\leqslant{i}\leqslant{n}} \:\:{n}\:{real}\:{number}\:\:{positifs}\:{wish}\:{verfy}\:\:\:\sum_{{i}=\mathrm{1}} ^{{i}={n}} \:{x}_{{i}} =\mathrm{1} \\ $$$${prove}\:{that}\:\:\:\sum_{\mathrm{1}\leqslant{i}\leqslant{n}} {x}_{{i}} ^{\mathrm{2}} \:\:\:\geqslant\:\:\frac{\mathrm{1}}{{n}}\:\:\:. \\ $$ Commented by…

lim-x-3-ln-1-5tan-4-x-x-1-cos-6-x-

Question Number 157314 by cortano last updated on 22/Oct/21 $$\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}\:\mathrm{ln}\:\left(\mathrm{1}+\mathrm{5tan}\:\frac{\mathrm{4}}{{x}}\right)}{{x}\:\left(\mathrm{1}−\mathrm{cos}\:\frac{\mathrm{6}}{{x}}\right)}\:=? \\ $$ Commented by john_santu last updated on 22/Oct/21 $$\:{let}\:\frac{\mathrm{1}}{{x}}\:=\:{u}\:{and}\:{u}\rightarrow\mathrm{0} \\ $$$$\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}{u}\:\mathrm{ln}\:\left(\mathrm{1}+\mathrm{5tan}\:\mathrm{4}{u}\right)}{\mathrm{1}−\mathrm{cos}\:\mathrm{6}{u}} \\…

let-put-x-n-1-1-n-x-with-x-gt-1-and-x-n-1-1-n-n-x-find-a-relation-between-x-and-x-

Question Number 26223 by abdo imad last updated on 22/Dec/17 $${let}\:{put}\:\xi\left({x}\right)=\:\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{\mathrm{1}}{{n}^{{x}} }\:\:{with}\:{x}>\mathrm{1} \\ $$$${and}\:\:\delta\left({x}\right)\:\:=\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{x}} }\:\:\:{find}\:{a}\:{relation} \\ $$$${between}\:\xi\left({x}\right)\:{and}\:\delta\left({x}\right). \\ $$ Commented…

find-the-value-of-n-1-1-n-1-n-2-n-3-in-terms-of-3-we-give-x-n-1-1-n-x-and-x-gt-1-zeta-function-of-Rieman-

Question Number 26222 by abdo imad last updated on 22/Dec/17 $${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right){n}^{\mathrm{3}} }\:{in}\:{terms}\:{of}\:\xi\left(\mathrm{3}\right) \\ $$$${we}\:{give}\:\xi\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{\mathrm{1}}{{n}^{{x}} }\:\:{and}\:{x}>\mathrm{1} \\ $$$$\left({zeta}\:{function}\:{of}\:{Rieman}\right) \\ $$ Commented by…

find-the-rsdius-of-convergence-for-theserie-n-0-x-n-2n-1-and-calculate-its-sum-s-x-find-the-value-of-n-0-1-2-n-2n-1-

Question Number 26192 by abdo imad last updated on 22/Dec/17 $${find}\:{the}\:{rsdius}\:{of}\:{convergence}\:{for}\:{theserie} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\propto} \:\frac{{x}^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:{and}\:{calculate}\:{its}\:{sum}\:{s}\left({x}\right) \\ $$$${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)}\:\:. \\ $$ Commented by…

let-put-S-n-k-1-k-n-1-k-k-find-S-n-in-terms-of-H-n-then-lim-n-gt-S-n-H-n-k-1-k-n-1-k-harmonic-serie-

Question Number 26132 by abdo imad last updated on 21/Dec/17 $${let}\:{put}\:{S}_{{n}} \:=\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}} \\ $$$${find}\:{S}_{{n}\:} {in}\:{terms}\:{of}\:\:{H}_{{n}} \:{then}\:{lim}_{{n}−>\propto} \:{S}_{{n}} \\ $$$${H}_{{n}} \:=\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \frac{\mathrm{1}}{{k}}\:\:\:\left(\:{harmonic}\:{serie}\right)…

answer-to26109-S-n-k-1-k-n-1-k-2-k-1-2-we-decompose-F-X-1-X-2-X-1-2-a-X-b-X-2-c-X-1-d-X-1-2-we-find-F-X-2-X-1-X-2-2-X-

Question Number 26121 by abdo imad last updated on 20/Dec/17 $${answer}\:{to}\mathrm{26109}\:\:\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{we}\:{decompose} \\ $$$${F}\left({X}\right)\:\:=\:\:\frac{\mathrm{1}}{{X}^{\mathrm{2}} \left({X}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\:\frac{{a}}{{X}}\:\:\:+\frac{{b}}{{X}^{\mathrm{2}^{} } }\:\:+\frac{{c}}{{X}+\mathrm{1}}\:\:+\frac{{d}}{\left({X}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{we}\:{find} \\ $$$${F}\left({X}\right)\:\:=\:\:\frac{−\mathrm{2}}{{X}}\:\:+\frac{\mathrm{1}}{{X}^{\mathrm{2}}…