6th Problem

Using the formula for the sum of geometric series, i.e.

$S_{\infty }=\frac{a}{1-r}\; for\; \left | r \right |< 1$ , in which $a$ is the first term of the series, and $r$ is the constant ratio.

Considering the first geometric series provided,

we have $S_{\infty }=\frac{\left ( \frac{a}{b} \right )}{1-\frac{1}{b}}=4$

$\Rightarrow \frac{\left ( \frac{a}{b} \right )}{\left ( \frac{b-1}{b} \right )}=4$

$\Rightarrow \frac{a}{b-1}=4$

$\Rightarrow a=4(b-1)\cdots \cdots \left \{ 1 \right \}.$

Now consider the second, using the same formula, getting

$S'_{\infty }=\frac{\left ( \frac{a}{a+b} \right )}{\left ( 1-\frac{1}{a+b} \right )}$

$=\frac{a}{a+b-1}$ .

By substituting $\left \{ 1 \right \}$ in, we have

$S'_{\infty }=\frac{4b-4}{5b-5}=\frac{4}{5}= \boxed{0.8}.$