Question

$\underset{n\to \infty }{lim}\frac{1+a+a^2+...+a^{n-1}}{1+b+b^2+...+b^{n-1}}=0$ if:

• A. $a=b$
• B. $a<b$
• C. $a>b$
• D. $a>1$

Answer 1

Answer: (B), (D)

The correct options are
B $a<b$
D $a>1$

$\underset{n\to \infty }{lim}\frac{1+a+a^2+...+a^{n-1}}{1+b+b^2+...+b^{n-1}}$

$=\underset{n\to \infty }{lim}\frac{1\cdot \frac{a^n-1}{a-1}}{1\cdot \frac{b^n-1}{b-1}}=\underset{n\to \infty }{lim}\frac{b-1}{a-1}\cdot \frac{a^n-1}{b^n-1}\frac{\infty }{\infty }$form

L Hospital

$=\underset{n\to \infty }{lim}\left(\frac{b-1}{a-1}\right)\cdot \frac{a^n\log a}{b^n\log b}=\frac{(b-1)\log a}{(a-1)\log b}\underset{n\to \infty }{lim}{\left(\frac{a}{b}\right)}^n$

$\underset{n\to \infty }{lim}{\left(\frac{a}{b}\right)}^n=0$ if $\frac{a}{b}<1$

$\Rightarrow a<b$

& $a>1$

because $\underset{n\to \infty }{lim}\frac{(a^n-1)}{(b^n-1)}=\frac{\infty }{\infty }$

if $a<1$

$\underset{n\to \infty }{lim}{a}^n={\left(\frac{1}{P}\right)}^n(P>1)$

$=\frac{1}{P^{\infty }}=\frac{1}{\infty }=0$