Question-167324

Question Number 167324 by mathlove last updated on 13/Mar/22

Answered by Rasheed.Sindhi last updated on 13/Mar/22

lim_(x→∞) (((1+(1/3)+(1/9)+...+(1/3^n ))/(1+(1/5)+(1/(25))+...+(1/5^n )))) ((1+(1/3)+(1/9)+...)/(1+(1/5)+(1/(25))+...))=((1/(1−(1/3)))/(1/(1−(1/5))))=((1/(2/3))/(1/(4/5)))=((3/2)/(5/4)) =(3/2)×(4/5)=(6/5)

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{9}}+…+\frac{\mathrm{1}}{\mathrm{3}^{{n}} }}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{25}}+…+\frac{\mathrm{1}}{\mathrm{5}^{{n}} }}\right) \\ $$$$\:\:\:\:\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{9}}+…}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{25}}+…}=\frac{\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}}{\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}}}=\frac{\frac{\mathrm{1}}{\mathrm{2}/\mathrm{3}}}{\frac{\mathrm{1}}{\mathrm{4}/\mathrm{5}}}=\frac{\frac{\mathrm{3}}{\mathrm{2}}}{\frac{\mathrm{5}}{\mathrm{4}}} \\ $$$$\:\:\:=\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{5}}=\frac{\mathrm{6}}{\mathrm{5}} \\ $$$$ \\ $$

Answered by amin96 last updated on 13/Mar/22

lim_(n→∞) (((Σ_(m=0) ^n (1/3^m ))/(Σ_(m=0) ^n (1/5^m ))))=((lim_(n→∞) Σ_(m=0) ^n ((1/3))^m )/(lim_(n→∞) Σ_(m=0) ^n ((1/5))^m ))=((lim_(n→∞) (((1−(1/3^(n+1) ))/(2/3))))/(lim_(n→∞) (((1−(1/5^(n+1) ))/(4/5)))))= =((3/2)/(5/4))=((12)/(10))=(6/5)

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\underset{\boldsymbol{\mathrm{m}}=\mathrm{0}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\frac{\mathrm{1}}{\mathrm{3}^{\boldsymbol{\mathrm{m}}} }}{\underset{\boldsymbol{\mathrm{m}}=\mathrm{0}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\frac{\mathrm{1}}{\mathrm{5}^{\boldsymbol{\mathrm{m}}} }}\right)=\frac{\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{\boldsymbol{\mathrm{m}}=\mathrm{0}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\boldsymbol{\mathrm{m}}} }{\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{\boldsymbol{\mathrm{m}}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{\boldsymbol{\mathrm{m}}} }=\frac{\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} }}{\frac{\mathrm{2}}{\mathrm{3}}}\right)}{\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}^{{n}+\mathrm{1}} }}{\frac{\mathrm{4}}{\mathrm{5}}}\right)}= \\ $$$$=\frac{\frac{\mathrm{3}}{\mathrm{2}}}{\frac{\mathrm{5}}{\mathrm{4}}}=\frac{\mathrm{12}}{\mathrm{10}}=\frac{\mathrm{6}}{\mathrm{5}} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply