Question Number 10671 by Saham last updated on 22/Feb/17 $$\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{5}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{n}\:−\:\mathrm{1}} \\ $$ Answered by FilupS last updated on 22/Feb/17 $${S}=\mathrm{5}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}^{{n}−\mathrm{1}} }…
Question Number 10670 by FilupS last updated on 22/Feb/17 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\zeta\left({s}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}^{−{s}} =\underset{{p}\in\mathbb{P}} {\overset{\infty} {\prod}}\left(\mathrm{1}−{p}^{−{s}} \right)^{−\mathrm{1}} \\ $$ Terms of Service Privacy Policy…
Question Number 10665 by FilupS last updated on 22/Feb/17 $${n}\mathrm{th}\:\mathrm{prime}\:=\:{p}_{{n}} \\ $$$${n}\mathrm{th}\:\mathrm{non}−\mathrm{prime}\:=\:{q}_{{n}} \\ $$$$\: \\ $$$$\mathrm{Determine}\:\mathrm{if}\:{q}_{{n}} >{p}_{{n}} \:\mathrm{for}\:\forall{n}\geqslant\mathrm{2} \\ $$ Commented by FilupS last updated…
Question Number 10664 by FilupS last updated on 22/Feb/17 $${S}=\underset{\underset{{n}\geqslant\mathrm{1}} {{n}\notin\mathbb{P}}} {\overset{\infty} {\sum}}{n} \\ $$$${Q}=\underset{\underset{{n}\geqslant\mathrm{1}} {{n}\in\mathbb{P}}} {\overset{\infty} {\sum}}{n} \\ $$$$\: \\ $$$$\mathrm{Prove}\:\mathrm{if}\:\mathrm{true}: \\ $$$${S}>{Q} \\…
Question Number 10662 by FilupS last updated on 22/Feb/17 $$\mathrm{determine}\:\mathrm{if}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{{s}} }+\frac{\mathrm{1}}{\mathrm{3}^{{s}} }+\frac{\mathrm{1}}{\mathrm{5}^{{s}} }+…\geqslant\frac{\mathrm{1}}{\mathrm{1}^{{s}} }+\frac{\mathrm{1}}{\mathrm{4}^{{s}} }+\frac{\mathrm{1}}{\mathrm{6}^{{s}} }+… \\ $$$$\mathrm{or}: \\ $$$$\underset{\underset{{n}\geqslant\mathrm{1}} {{n}\in\mathbb{P}}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{s}}…
Question Number 10627 by pan123 last updated on 20/Feb/17 $$\mathrm{60}^{\mathrm{15}} ×\mathrm{30}^{\mathrm{8}} ×\mathrm{35}^{\mathrm{6}} ×\mathrm{60}^{\mathrm{15}} =? \\ $$ Answered by mrW1 last updated on 20/Feb/17 $$\mathrm{60}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}×\mathrm{5}…
Question Number 141694 by Dwaipayan Shikari last updated on 22/May/21 $${log}\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{10}}\mathrm{9}{e}^{\gamma} \right)=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{2}}\left(\frac{\mathrm{1}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} }{\mathrm{10}^{\mathrm{2}} }\right)−\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{3}}\:\left(\frac{\mathrm{1}^{\mathrm{3}} +\mathrm{9}^{\mathrm{3}} }{\mathrm{10}^{\mathrm{3}} }\:\right)+\frac{\zeta\left(\mathrm{4}\right)}{\mathrm{4}}\left(\frac{\mathrm{1}^{\mathrm{4}} +\mathrm{9}^{\mathrm{4}} }{\mathrm{10}^{\mathrm{4}} }\right)−… \\ $$$$\gamma={Euler}\:{Mascheroni}\:{Constant} \\ $$…
Question Number 10577 by pan123 last updated on 19/Feb/17 $$\mathrm{5}^{\mathrm{71}} +\mathrm{5}^{\mathrm{72}} +\mathrm{5}^{\mathrm{73}} =? \\ $$ Answered by ridwan balatif last updated on 19/Feb/17 $$\mathrm{5}^{\mathrm{71}} +\mathrm{5}^{\mathrm{72}}…
Question Number 10539 by malwaan last updated on 17/Feb/17 $${prove}\:{that} \\ $$$$\sqrt{\mathrm{2}\:+\overset{\mathrm{3}} {\:}\sqrt{\mathrm{3}\:+…+\overset{\mathrm{1993}} {\:}\sqrt{\mathrm{1993}}}}\:<\mathrm{2} \\ $$ Answered by mrW1 last updated on 17/Feb/17 $${function}\:{y}=\left({x}+{a}\right)^{\frac{\mathrm{1}}{{x}}} \:{with}\:{a}>\mathrm{1}\:{is}…
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