Esmeralda Lane

2022-07-07

I got stuck trying to solve this limit.
$\underset{n\to \mathrm{\infty }}{lim}\frac{1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{{2}^{n}}}{1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{{3}^{n}}}$

behk0

We see, that $\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{{2}^{n}}\right)$ and $\left(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{{3}^{n}}\right)$ are the geometric series.
Therefore,
$\underset{n\to \mathrm{\infty }}{lim}\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{{2}^{n}}\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^{n}=\frac{1}{1-\frac{1}{2}}=2,$
$\underset{n\to \mathrm{\infty }}{lim}\left(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{{3}^{n}}\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{3}\right)}^{n}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}$
We will have
$\underset{n\to \mathrm{\infty }}{lim}\frac{\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{{2}^{n}}\right)}{\left(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{{3}^{n}}\right)}=\frac{2}{\frac{3}{2}}=\frac{4}{3}.$

Do you have a similar question?