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Question Number 5519 by FilupSmith last updated on 17/May/16

P=Π_(i=1) ^n (a^i /b^i ) a≠b, b≠0 P=??

$${P}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{{a}^{{i}} }{{b}^{{i}} }\:\:\:\:\:\:\:\:{a}\neq{b},\:{b}\neq\mathrm{0} \\ $$$${P}=?? \\ $$

Commented by Yozzii last updated on 17/May/16

((a×a^2 ×a^3 ×...×a^n )/(b×b^2 ×b^3 ×...×b^n ))=(a^(1+2+3+...+n) /b^(1+2+3+...+n) )=(a^(0.5n(n+1)) /b^(0.5n(n+1)) )=((a/b))^(0.5n(n+1)) P=(a/b)^(0.5n(n+1))

$$\frac{{a}×{a}^{\mathrm{2}} ×{a}^{\mathrm{3}} ×...×{a}^{{n}} }{{b}×{b}^{\mathrm{2}} ×{b}^{\mathrm{3}} ×...×{b}^{{n}} }=\frac{{a}^{\mathrm{1}+\mathrm{2}+\mathrm{3}+...+{n}} }{{b}^{\mathrm{1}+\mathrm{2}+\mathrm{3}+...+{n}} }=\frac{{a}^{\mathrm{0}.\mathrm{5}{n}\left({n}+\mathrm{1}\right)} }{{b}^{\mathrm{0}.\mathrm{5}{n}\left({n}+\mathrm{1}\right)} }=\left(\frac{{a}}{{b}}\right)^{\mathrm{0}.\mathrm{5}{n}\left({n}+\mathrm{1}\right)} \\ $$$${P}=\left({a}/{b}\right)^{\mathrm{0}.\mathrm{5}{n}\left({n}+\mathrm{1}\right)} \\ $$

Commented by FilupSmith last updated on 17/May/16

P=((a∙a^2 ∙a^3 ∙...∙a^n )/(b∙b^2 ∙b^3 ∙...∙a^n )) P=(a^((1/2)n(n+1)) /b^((1/2)n(n+1)) ) ∴P=((a/b))^((n(n+1))/2) What if n→∞??

$${P}=\frac{{a}\centerdot{a}^{\mathrm{2}} \centerdot{a}^{\mathrm{3}} \centerdot...\centerdot{a}^{{n}} }{{b}\centerdot{b}^{\mathrm{2}} \centerdot{b}^{\mathrm{3}} \centerdot...\centerdot{a}^{{n}} } \\ $$$${P}=\frac{{a}^{\frac{\mathrm{1}}{\mathrm{2}}{n}\left({n}+\mathrm{1}\right)} }{{b}^{\frac{\mathrm{1}}{\mathrm{2}}{n}\left({n}+\mathrm{1}\right)} } \\ $$$$\therefore{P}=\left(\frac{{a}}{{b}}\right)^{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}} \\ $$$$\mathrm{What}\:\mathrm{if}\:{n}\rightarrow\infty?? \\ $$

Commented by Yozzii last updated on 17/May/16

lim_(n→∞) (a/b)^(0.5n(n+1)) = { ((divergent if a≥−b)),((0 if −1<(a/b)<1)),((divergent if a>b)) :}

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left({a}/{b}\right)^{\mathrm{0}.\mathrm{5}{n}\left({n}+\mathrm{1}\right)} =\begin{cases}{{divergent}\:\:\:\:\:{if}\:{a}\geqslant−{b}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:−\mathrm{1}<\left({a}/{b}\right)<\mathrm{1}}\\{{divergent}\:\:\:\:\:{if}\:{a}>{b}}\end{cases} \\ $$

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