Question Number 127085 by mnjuly1970 last updated on 26/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:{calculus}… \\ $$$$\:{prove}\:{that}:: \\ $$$$\:\:\:{lim}_{\:{n}\rightarrow\infty} \left(\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{k}}{{n}}\right)^{{n}} \right)=\frac{{e}}{{e}−\mathrm{1}} \\ $$$$ \\ $$ Commented by mnjuly1970…
Question Number 192617 by mechanics last updated on 23/May/23 Answered by a.lgnaoui last updated on 23/May/23 $$\mathrm{0} \\ $$ Answered by Skabetix last updated on…
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Question Number 127080 by Dwaipayan Shikari last updated on 26/Dec/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{e}^{−\phi{n}} +\frac{{e}^{\mathrm{2}\pi{n}} −{e}^{−\mathrm{2}\phi{n}} \:}{\mathrm{2}{e}^{−\phi{n}} +\frac{{e}^{\mathrm{2}\pi{n}} −{e}^{−\mathrm{2}\phi{n}} }{\mathrm{2}{e}^{−\phi{n}} +\frac{{e}^{\mathrm{2}\pi{n}} −{e}^{−\mathrm{2}\phi{n}} }{\mathrm{2}{e}^{−\mathrm{2}\phi{n}} …}}}} \\ $$…
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Question Number 127081 by slahadjb last updated on 26/Dec/20 $${Solve}\:{in}\:\mathbb{C} \\ $$$$\left(\frac{\mathrm{1}+{iz}}{\mathrm{1}−{iz}}\right)^{{n}} ={e}_{} ^{{i}\theta_{{n}} } \\ $$$$\theta_{{n}} \:\in\:\mathbb{R} \\ $$$${n}\:\in\:\mathbb{N} \\ $$ Answered by Dwaipayan…
Question Number 127078 by amns last updated on 26/Dec/20 $$\frac{\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}}{\frac{\frac{\frac{\mathrm{1}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}{\mathrm{10}}}{\mathrm{3}}}{\mathrm{2}}}\:=\:? \\ $$ Commented by hknkrc46 last updated on 26/Dec/20 $$\left(\frac{\mathrm{1}}{\mathrm{2}}\::\:\sqrt{\mathrm{2}}\right)\::\:\left[\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\::\:\mathrm{10}\::\:\mathrm{3}\::\:\mathrm{2}\right] \\ $$$$=\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:\centerdot\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\::\:\left[\frac{\sqrt{\mathrm{2}}\:+\:\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\centerdot\:\frac{\mathrm{1}}{\mathrm{10}}\:\centerdot\:\frac{\mathrm{1}}{\mathrm{3}}\:\centerdot\:\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\centerdot\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\centerdot\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}\:+\:\mathrm{1}}\:\centerdot\:\frac{\mathrm{10}}{\mathrm{1}}\:\centerdot\:\frac{\mathrm{3}}{\mathrm{1}}\:\centerdot\:\frac{\mathrm{2}}{\mathrm{1}} \\…
Question Number 192612 by York12 last updated on 22/May/23 $$\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{z}}\:=\:\mathrm{1}\: \\ $$$${find}\:{the}\:{minimum}\:{value}\:{of}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \\ $$ Commented by Frix last updated on 23/May/23 $$\mathrm{Yes}\:\mathrm{of}\:\mathrm{course}! \\…
Question Number 192608 by York12 last updated on 22/May/23 $${let}\:{k}\:{be}\:{natural}\:{number}.\:{defined}\:{s}_{{k}} \:{as}\:{the} \\ $$$${sum}\:{of}\:{the}\:{infinite}\:{series}\:{s}_{{k}} =\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{0}} }\:+\:\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{1}} }\:+\:\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} }\:+… \\ $$$${find}\:{the}\:{value}\:{of}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left[\frac{{s}_{{k}} }{\mathrm{2}^{{k}−\mathrm{1}}…
Question Number 61537 by aanur last updated on 04/Jun/19 Commented by aanur last updated on 04/Jun/19 $${sir}\:{could}\:{you}\:{help}\:{me} \\ $$ Commented by math1967 last updated on…