Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

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

$$\underset{x\to \mathrm{\infty }}{\lim }\left(\frac{1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{3^n}}{1+\frac{1}{5}+\frac{1}{25}+...+\frac{1}{5^n}}\right)$$$$\frac{1+\frac{1}{3}+\frac{1}{9}+...}{1+\frac{1}{5}+\frac{1}{25}+...}=\frac{\frac{1}{1-\frac{1}{3}}}{\frac{1}{1-\frac{1}{5}}}=\frac{\frac{1}{2/3}}{\frac{1}{4/5}}=\frac{\frac{3}{2}}{\frac{5}{4}}$$$$=\frac{3}{2}\times \frac{4}{5}=\frac{6}{5}$$$$$$

Answered by amin96 last updated on 13/Mar/22

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com