Let $a$ and $b$ be real numbers such that
\[\frac{a}{b} + \frac{a}{b^2} + \frac{a}{b^3} + \dots = 4.\]Find
\[\frac{a}{a + b} + \frac{a}{(a + b)^2} + \frac{a}{(a + b)^3} + \dotsb.\]
From the formula for an infinite geometric series,
\[\frac{a/b}{1 - 1/b} = 4.\]Then $\frac{a}{b - 1} = 4,$ so $a = 4(b - 1).$

Again from the formula,
\begin{align*}
\frac{a}{a + b} + \frac{a}{(a + b)^2} + \frac{a}{(a + b)^3} + \dotsb &= \frac{a/(a + b)}{1 - 1/(a + b)} \\
&= \frac{a}{a + b - 1} \\
&= \frac{4(b - 1)}{4(b - 1) + (b - 1)} \\
&= \frac{4(b - 1)}{5(b - 1)} = \boxed{\frac{4}{5}}.
\end{align*}
\frac{4}{5}
Level 4
Intermediate Algebra
Problem: Let $a$ and $b$ be real numbers such that
\[\frac{a}{b} + \frac{a}{b^2} + \frac{a}{b^3} + \dots = 4.\]Find
\[\frac{a}{a + b} + \frac{a}{(a + b)^2} + \frac{a}{(a + b)^3} + \dotsb.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we first recognize that the series given in the problem is a geometric series. The series can be expressed as:

\[\frac{a}{b} + \frac{a}{b^2} + \frac{a}{b^3} + \dots = \frac{a/b}{1 - 1/b} = \frac{a/b}{(b-1)/b} = \frac{a}{b-1}.\]

Setting this equal to 4 gives us the equation:

\[\frac{a}{b-1} = 4.\]

From this, we can derive that:

\[a = 4(b - 1) = 4b - 4.\]

Next, we need to evaluate the second series:

\[\frac{a}{a + b} + \frac{a}{(a + b)^2} + \frac{a}{(a + b)^3} + \dotsb.\]

This series is also a geometric series, and we can express it as:

\[\frac{a}{a + b} \cdot \frac{1}{1 - \frac{1}{a + b}} = \frac{a/(a + b)}{1 - 1/(a + b)} = \frac{a/(a + b)}{(a + b - 1)/(a + b)} = \frac{a}{(a + b - 1)}.\]

Now, we need to find \(a + b\). We substitute \(a = 4b - 4\) into \(a + b\):

\[a + b = (4b - 4) + b = 5b - 4.\]

Next, we substitute \(a\) and \(a + b\) into the expression for the second series:

\[\frac{a}{a + b - 1} = \frac{4b - 4}{(5b - 4) - 1} = \frac{4b - 4}{5b - 5} = \frac{4(b - 1)}{5(b - 1)}.\]

Assuming \(b \neq 1\), we can simplify this to:

\[\frac{4}{5}.\]

Thus, the final answer is \(\frac{4}{5}\).<|END_OF_LATENT|> The final answer is $\frac{4}{5}$.