Question Number 223049 by wewji12 last updated on 13/Jul/25 $$\mathrm{we}\:\mathrm{already}\:\mathrm{know}\:\:\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{h}}=\infty\:\mathrm{and}\:\underset{{p}\in\mathbb{P}} {\sum}\:\frac{\mathrm{1}}{{p}}=\infty \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{Harmonic}\:\mathrm{Series} \\ $$$$\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{h}}=\underset{{p}\in\mathbb{P}} {\prod}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{p}}+\left(\frac{\mathrm{1}}{{p}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{p}}\right)^{\mathrm{3}} +…\right)=\:\underset{{p}\in\mathbb{P}} {\prod}\:\frac{\mathrm{1}}{\mathrm{1}−{p}^{−\mathrm{1}} } \\…
Question Number 223000 by MrGaster last updated on 12/Jul/25 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\mathrm{1}}{{e}}\left(\frac{{n}+\mathrm{1}}{{n}}\right)^{{n}} \right)^{\left(−\mathrm{1}\right)^{{n}} } =\frac{{e}\centerdot\sqrt{\pi}\centerdot\sqrt[{\mathrm{6}}]{\mathrm{2}}}{{A}^{\mathrm{6}} } \\ $$ Commented by MathematicalUser2357 last updated on 22/Jul/25…
Question Number 222954 by klipto last updated on 11/Jul/25 $$ \\ $$ Commented by klipto last updated on 11/Jul/25 Commented by gabthemathguy25 last updated on…
Question Number 222845 by Nicholas666 last updated on 09/Jul/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Evaluate}; \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:\left(−\mathrm{1}\right)^{{k}} \:\begin{pmatrix}{\mathrm{2}{n}\:−\:{k}}\\{\:\:\:\:\:\:\:\:{k}}\end{pmatrix} \\ $$$$ \\ $$ Terms of Service Privacy…
Question Number 222828 by wewji12 last updated on 09/Jul/25 $$\mathrm{vector}\:\mathrm{field}\:\:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}};\mathbb{R}^{\mathrm{3}} \rightarrow\mathbb{R}^{\mathrm{3}} \:,\:{F}_{{h}} \in\mathcal{C}^{\omega} \\ $$$$\mathrm{and}\:\mathrm{Let}'\mathrm{s}\:\mathrm{define}\:\mathrm{as}\:\overset{\rightarrow} {\boldsymbol{\mathrm{A}}}=\overset{\rightarrow} {\bigtriangledown}×\overset{\rightarrow} {\boldsymbol{\mathrm{F}}} \\ $$$$\mathrm{can}\:\mathrm{we}\:\mathrm{find}\:\mathrm{vector}\:\mathrm{field}\:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}…..??? \\ $$$$\mathrm{Curl}\:\mathrm{and}\:\mathrm{Divergence}\:\:\mathrm{inverse}\:\mathrm{operator}\:\mathrm{dose}\:\mathrm{exist}?? \\…
Question Number 222711 by sonukgindia last updated on 05/Jul/25 Answered by gabthemathguy25 last updated on 05/Jul/25 $$\left({ABC}\right)_{\mathrm{16}} =\mathrm{10}×\mathrm{16}^{\mathrm{2}} +\mathrm{11}×\mathrm{16}^{\mathrm{1}} +\mathrm{12}×\mathrm{16}^{\mathrm{0}} \\ $$$$\Rightarrow\mathrm{10}×\mathrm{256}+\mathrm{11}×\mathrm{16}+\mathrm{12}=\mathrm{2560}+\mathrm{176}+\mathrm{12}=\mathrm{2748} \\ $$$$\mathrm{its}\:\mathrm{1},\:\mathrm{2748}. \\…
Question Number 222662 by MrGaster last updated on 04/Jul/25 Commented by MrGaster last updated on 04/Jul/25 $$\exists\bigtriangleup{ABC},\angle{B}=\mathrm{90}°,{BA}={AC},{BD}=\mathrm{3}\wedge\angle{CED}=\mathrm{45}°,{CE}=\sqrt{\mathrm{2}}{AE},{CE}=? \\ $$ Commented by mr W last updated…
Question Number 222619 by ahmed2025 last updated on 01/Jul/25 Answered by wewji12 last updated on 02/Jul/25 $$\int\:\:\frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:\:\mathrm{d}{x}=−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\left(\mathrm{2}−\mathrm{3}{x}\right)+{C} \\ $$$$\int_{−\mathrm{1}} ^{\:\:\mathrm{2}} \:\frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:\mathrm{d}{x}=\int_{\:−\mathrm{1}} ^{\frac{\mathrm{2}}{\mathrm{3}}} \frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:\mathrm{d}{x}\:+\int_{\frac{\mathrm{2}}{\mathrm{3}}} ^{\:\mathrm{2}} \:\frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:\mathrm{d}{x}…
Question Number 222599 by MrGaster last updated on 01/Jul/25 $$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }=\frac{\pi^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} \left(\mathrm{2}{n}\right)!}\mid{E}_{\mathrm{2}{m}} \mid \\ $$ Commented by MathematicalUser2357 last updated on…
Question Number 222532 by MrGaster last updated on 29/Jun/25 $$\mathrm{Prove}: \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}{n}} }\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}^{\mathrm{2}} \frac{\mathscr{K}_{{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{192}}{\pi}\mathfrak{I}\mathrm{Li}_{\mathrm{4}} \left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)+\frac{\mathrm{16}}{\pi}\mathfrak{I}\mathrm{Li}_{\mathrm{3}} \left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)\mathrm{ln}\left(\mathrm{2}\right)+\frac{\mathrm{15}\pi^{\mathrm{2}} }{\mathrm{8}}\mathrm{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)^{\mathrm{3}} −\frac{\mathrm{148}}{\pi}\beta\left(\mathrm{4}\right),\mathscr{K}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\ldots+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}} \\ $$…