Question

Find the following limit.
$\underset{n\to \infty }{lim}(1-\frac{1}{7}+\frac{1}{49}-...+\frac{(-1{)}^{n-1}}{7^{n-2}}).$

Answer 1

$=\underset{n\to \infty }{lim}(1-\frac{1}{7}+\frac{1}{49}.....+\frac{(1{)}^{n-1}}{7^{n-2}})$

this is a G.P. with first term $=1$

and common ration $=\frac{-1}{7}$

So sum of $n$ terms of G.P.

$S_n=1-\frac{1}{7}+\frac{1}{49}+....+\frac{(1{)}^{n-1}}{7^{n-2}}$

$S_n=\frac{1(1-{\left(\frac{-1}{7}\right)}^n)}{1-\left(\frac{-1}{7}\right)}[S_n=\frac{a(1-r{)}^n)}{1-r}]$

So $\underset{n\to \infty }{lim}\frac{1-{(-\frac{1}{7})}^n}{1+\frac{1}{7}}$

$=\underset{n\to \infty }{lim}\frac{1-{(-\frac{1}{7})}^n}{\frac{8}{7}}$

$=\frac{1-{(-\frac{1}{7})}^{\infty }}{\frac{8}{7}}=\frac{7}{8}$.