Suppose that $a$ is a positive integer with $a>1$. Determine a closed form expression, in terms of $a$, equal to

$$
1+\frac{3}{a}+\frac{5}{a^{2}}+\frac{7}{a^{3}}+\cdots
$$

(The infinite sum includes exactly the fractions of the form $\frac{2 k-1}{a^{k-1}}$ for each positive integer $k$.)
Suppose that $S=1+\frac{3}{a}+\frac{5}{a^{2}}+\frac{7}{a^{3}}+\cdots$.

Then $a S=a+3+\frac{5}{a}+\frac{7}{a^{2}}+\frac{9}{a^{3}}+\cdots$.

Therefore $a S-S=a+2+\frac{2}{a}+\frac{2}{a^{2}}+\frac{2}{a^{3}}+\cdots$.

Since $a>1$, then $2+\frac{2}{a}+\frac{2}{a^{2}}+\frac{2}{a^{3}}+\cdots$ is an infinite geometric series with common ratio $\frac{1}{a}<1$.

Therefore, $2+\frac{2}{a}+\frac{2}{a^{2}}+\frac{2}{a^{3}}+\cdots=\frac{2}{1-\frac{1}{a}}=\frac{2 a}{a-1}$.

Substituting, we obtain $a S-S=a+\frac{2 a}{a-1}$ and so $(a-1) S=\frac{\left(a^{2}-a\right)+2 a}{a-1}$ which gives $(a-1) S=\frac{a^{2}+a}{a-1}$ or $S=\frac{a^{2}+a}{(a-1)^{2}}$.

ANSWER: $\frac{a^{2}+a}{(a-1)^{2}}$