Question Number 12828 by 786786AM last updated on 03/May/17
$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{infinite}\:\mathrm{product} \\ $$$$\sqrt{\mathrm{3}}\:\centerdot\:\sqrt[{\mathrm{4}}]{\mathrm{9}}\:\centerdot\:\sqrt[{\mathrm{8}}]{\mathrm{27}}\:\centerdot\:\sqrt[{\mathrm{16}}]{\mathrm{81}}\:...\mathrm{to}\:\infty\:\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\_\_\_\_. \\ $$
Answered by prakash jain last updated on 04/May/17
$$\mathrm{3}^{\mathrm{1}/\mathrm{2}} \centerdot\mathrm{3}^{\mathrm{2}/\mathrm{4}} \centerdot\mathrm{3}^{\mathrm{3}/\mathrm{8}} ...=\mathrm{3}^{\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{2}^{\mathrm{4}} }+...\right)} =\mathrm{3}^{{S}} \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{2}^{\mathrm{4}} }+...\:\:\left({i}\right) \\ $$$$\frac{{S}}{\mathrm{2}}=\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{4}} }+....\:\:\left({ii}\right) \\ $$$${subtracting}\:\left({ii}\right)\:{from}\:\left({i}\right) \\ $$$$\frac{{S}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }+...=\frac{\mathrm{1}/\mathrm{2}}{\mathrm{1}−\mathrm{1}/\mathrm{2}}=\mathrm{1} \\ $$$${S}=\mathrm{2} \\ $$$$\mathrm{3}^{{S}} =\mathrm{3}^{\mathrm{2}} =\mathrm{9} \\ $$
Commented by mrW1 last updated on 04/May/17
$${very}\:{nice}\:{trick}! \\ $$