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Author: Tinku Tara

Need-help-solving-0-x-x-dx-My-current-working-x-x-e-xlnx-e-xlnx-1-xln-x-x-2-ln-x-2-2-x-3-ln-x-3-3-n-0-1-n-x-n-ln-n-x-n-0-x-

Question Number 6246 by FilupSmith last updated on 20/Jun/16 $$\mathrm{Need}\:\mathrm{help}\:\mathrm{solving}\:\int_{\mathrm{0}} ^{\:\infty} {x}^{−{x}} {dx} \\ $$$$\mathrm{My}\:\mathrm{current}\:\mathrm{working}: \\ $$$${x}^{−{x}} ={e}^{−{x}\mathrm{ln}{x}} \\ $$$${e}^{−{x}\mathrm{ln}{x}} =\mathrm{1}−{x}\mathrm{ln}\left({x}\right)+\frac{{x}^{\mathrm{2}} \mathrm{ln}\left({x}\right)^{\mathrm{2}} }{\mathrm{2}!}−\frac{{x}^{\mathrm{3}} \mathrm{ln}\left({x}\right)^{\mathrm{3}} }{\mathrm{3}!}+……

Question-71776

Question Number 71776 by TawaTawa last updated on 19/Oct/19 Answered by tw000001 last updated on 22/Oct/19 $$\mathrm{Use}\:\mathrm{Harmonic}\:\mathrm{series}\:\mathrm{to}\:\mathrm{solve}. \\ $$$${A}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{\mathrm{2015}}−\frac{\mathrm{1}}{\mathrm{2016}} \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{\mathrm{2016}}\right)−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{6}}+…+\frac{\mathrm{1}}{\mathrm{2016}}\right) \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{\mathrm{2016}}\right)−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{\mathrm{1008}}\right) \\ $$$$={H}_{\mathrm{2016}}…

lim-n-n-3-n-n-n-3-n-

Question Number 71777 by Henri Boucatchou last updated on 19/Oct/19 $$\:\:\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{lim}}}\:\left(\frac{\boldsymbol{{n}}!\:+\:\mathrm{3}^{\boldsymbol{{n}}} }{\boldsymbol{{n}}^{\boldsymbol{{n}}} \:+\:\mathrm{3}^{\boldsymbol{{n}}} }\right)\:=\:? \\ $$ Commented by mathmax by abdo last updated on…

Question-6238

Question Number 6238 by sanusihammed last updated on 19/Jun/16 Answered by Yozzii last updated on 20/Jun/16 $${Let}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+{n}^{−\mathrm{1}} \right)^{−{n}^{\mathrm{2}} } ={l}. \\ $$$${Let}\:{u}={n}^{−\mathrm{1}} \Rightarrow{l}=\underset{{u}\rightarrow\mathrm{0}^{+} }…

advanced-calculus-prove-that-n-0-1-n-4n-2-1-1-4-2-picsch-pi-2-

Question Number 137307 by mnjuly1970 last updated on 31/Mar/21 $$\:\:\:\:\:\:\:\:…..{advanced}\:\:\:\:{calculus}….. \\ $$$$\:\:\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}^{\mathrm{2}} +\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}+\pi{csch}\left(\frac{\pi}{\mathrm{2}}\right)\right) \\ $$$$\:\:\:\:\:\:………………………. \\ $$ Answered by Dwaipayan…