Question Number 5519 by FilupSmith last updated on 17/May/16
$${P}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{{a}^{{i}} }{{b}^{{i}} }\:\:\:\:\:\:\:\:{a}\neq{b},\:{b}\neq\mathrm{0} \\ $$$${P}=?? \\ $$
Commented by Yozzii last updated on 17/May/16
$$\frac{{a}×{a}^{\mathrm{2}} ×{a}^{\mathrm{3}} ×…×{a}^{{n}} }{{b}×{b}^{\mathrm{2}} ×{b}^{\mathrm{3}} ×…×{b}^{{n}} }=\frac{{a}^{\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{n}} }{{b}^{\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{n}} }=\frac{{a}^{\mathrm{0}.\mathrm{5}{n}\left({n}+\mathrm{1}\right)} }{{b}^{\mathrm{0}.\mathrm{5}{n}\left({n}+\mathrm{1}\right)} }=\left(\frac{{a}}{{b}}\right)^{\mathrm{0}.\mathrm{5}{n}\left({n}+\mathrm{1}\right)} \\ $$$${P}=\left({a}/{b}\right)^{\mathrm{0}.\mathrm{5}{n}\left({n}+\mathrm{1}\right)} \\ $$
Commented by FilupSmith last updated on 17/May/16
$${P}=\frac{{a}\centerdot{a}^{\mathrm{2}} \centerdot{a}^{\mathrm{3}} \centerdot…\centerdot{a}^{{n}} }{{b}\centerdot{b}^{\mathrm{2}} \centerdot{b}^{\mathrm{3}} \centerdot…\centerdot{a}^{{n}} } \\ $$$${P}=\frac{{a}^{\frac{\mathrm{1}}{\mathrm{2}}{n}\left({n}+\mathrm{1}\right)} }{{b}^{\frac{\mathrm{1}}{\mathrm{2}}{n}\left({n}+\mathrm{1}\right)} } \\ $$$$\therefore{P}=\left(\frac{{a}}{{b}}\right)^{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}} \\ $$$$\mathrm{What}\:\mathrm{if}\:{n}\rightarrow\infty?? \\ $$
Commented by Yozzii last updated on 17/May/16
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left({a}/{b}\right)^{\mathrm{0}.\mathrm{5}{n}\left({n}+\mathrm{1}\right)} =\begin{cases}{{divergent}\:\:\:\:\:{if}\:{a}\geqslant−{b}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:−\mathrm{1}<\left({a}/{b}\right)<\mathrm{1}}\\{{divergent}\:\:\:\:\:{if}\:{a}>{b}}\end{cases} \\ $$