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Question Number 137376 by benjo_mathlover last updated on 02/Apr/21

1+(1/3)+(1/5)+(1/9)+(1/(15))+(1/(25))+...+(1/(45))+(1/(75))+... =?

$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{25}}+...+\frac{\mathrm{1}}{\mathrm{45}}+\frac{\mathrm{1}}{\mathrm{75}}+...\:=? \\ $$

Answered by EDWIN88 last updated on 02/Apr/21

determinant ((,(1/3^0 ),(1/3^1 ),(1/3^2 ),(1/3^3 ),(...),(1/3^n ),Σ_(n=0) ^∞ ),((1/5^0 ),(1/(5^0 .3^0 )),(1/(5^0 .3^1 )),(1/(5^0 .3^2 )),(1/(5^0 .3^3 )),(...),(1/(5^0 .3^n )),((1/5^0 )Σ_(n=0) ^∞ (1/3^n ))),((1/5^1 ),(1/(5^1 .3^0 )),(1/(5^1 .3^1 )),(1/(5^1 .3^2 )),(1/(5^1 .3^3 )),(...),(1/(5^1 .3^n )),((1/5^1 )Σ_(n=0) ^∞ (1/3^n ))),((...),(...),(...),(...),(...),(...),(...),(...)),((1/5^m ),(1/(5^m .3^0 )),(1/(5^m .3^1 )),(1/(5^m .3^2 )),(1/(5^m .3^3 )),(...),(1/(5^m .3^n )),((1/5^m ) Σ_(n=0) ^∞ (1/3^n )))) so we get : ((1/5^0 )+(1/5^1 )+(1/5^2 )+...+(1/5^m )).Σ_(n=0) ^∞ (1/3^n ) = (Σ_(m=0) ^∞ (1/5^m )).(Σ_(n=0) ^∞ (1/3^n )) = (1/(1−(1/5))) .(1/(1−(1/3))) = ((15)/8)

$$\begin{array}{|c|c|c|c|c|}{}&\hline{\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{0}} }}&\hline{\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{1}} }}&\hline{\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }}&\hline{\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }}&\hline{...}&\hline{\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{n}} }}&\hline{\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}}\\{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{0}} }}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{0}} .\mathrm{3}^{\mathrm{0}} }}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{0}} .\mathrm{3}^{\mathrm{1}} }}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{0}} .\mathrm{3}^{\mathrm{2}} }}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{0}} .\mathrm{3}^{\mathrm{3}} }}&\hline{...}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{0}} .\mathrm{3}^{\mathrm{n}} }}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{0}} }\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{n}} }}\\{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{1}} }}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{1}} .\mathrm{3}^{\mathrm{0}} }}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{1}} .\mathrm{3}^{\mathrm{1}} }}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{1}} .\mathrm{3}^{\mathrm{2}} }}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{1}} .\mathrm{3}^{\mathrm{3}} }}&\hline{...}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{1}} .\mathrm{3}^{\mathrm{n}} }}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{1}} }\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{n}} }}\\{...}&\hline{...}&\hline{...}&\hline{...}&\hline{...}&\hline{...}&\hline{...}&\hline{...}\\{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{m}} }}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{m}} .\mathrm{3}^{\mathrm{0}} }}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{m}} .\mathrm{3}^{\mathrm{1}} }}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{m}} .\mathrm{3}^{\mathrm{2}} }}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{m}} .\mathrm{3}^{\mathrm{3}} }}&\hline{...}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{m}} .\mathrm{3}^{\mathrm{n}} }}&\hline{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{m}} }\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{n}} }}\\\hline\end{array} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{get}\::\:\left(\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{0}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }+...+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{m}} }\right).\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{n}} }\:= \\ $$$$\left(\underset{\mathrm{m}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{m}} }\right).\left(\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{n}} }\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}}\:.\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}\:=\:\frac{\mathrm{15}}{\mathrm{8}} \\ $$

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