Question Number 167955 by peter frank last updated on 30/Mar/22
Answered by JDamian last updated on 30/Mar/22
$$\mathrm{6}^{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\centerdot\centerdot\centerdot} =\mathrm{6}^{\mathrm{1}} =\mathrm{6} \\ $$
Answered by alephzero last updated on 30/Mar/22
$$\mathrm{6}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}+...} \:=\:\mathrm{exp}_{\mathrm{6}} \left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right) \\ $$$${S}\:=\:\frac{{a}_{\mathrm{1}} }{\mathrm{1}−{q}}\:\:\:\:\:{a}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\wedge\:{q}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:{S}\:=\:\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)}\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{6}^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{6}^{\frac{\mathrm{1}}{\mathrm{4}}} \mathrm{6}^{\frac{\mathrm{1}}{\mathrm{8}}} ...\:=\:\mathrm{6}^{\mathrm{1}} \:=\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\blacksquare\:{QED} \\ $$
Commented by peter frank last updated on 30/Mar/22
$$\mathrm{thank}\:\mathrm{you} \\ $$