Question

$\underset{n\to \infty }{lim}\frac{1+5+5^2+...+5^n}{1-{25}^n}$

• A. $0$
• B. $-1$
• C. $1$
• D. $\infty$

Answer 1

The correct option is A $0$

Let, $L=\underset{n\to \infty }{lim}\frac{1+5+5^2+...+5^n}{1-{25}^n}=\underset{n\to \infty }{lim}\frac{\frac{1-5^{n+1}}{1-5}}{1-{25}^n}$
$=-\frac{1}{4}\underset{n\to \infty }{lim}\frac{1-5^{n+1}}{1-{25}^n}$
Now Using l-hospital rule,
$L=-\frac{1}{4}\underset{n\to \infty }{lim}\frac{-5^{n+1}\log 5}{-{25}^n\log 25}=-\frac{1}{8}\underset{n\to \infty }{lim}\frac{-5^{n+1}}{-5^{2n}}=-\frac{1}{8}\underset{n\to \infty }{lim}\frac{1}{5^{n-1}}=0$