a=b+b^2+b^3+b^4...Value of b lies between -1/2 and 1/2. GP a-a^2+a^3-a^4... is to be found.

Question

Based upon this info, we are to find $a$-$a^2$+$a^3$-$a^4$... I could find that $a$=$b$/$(1-b)$ as it is infinite geometric progression. I tried doing mathematical induction but was stuck with a very tedious answer. Is there any better method for the same? Answer is b.

Answer 1

Since $a=\frac b{1-b}$,$$a-a^2+a^3-a^4+\cdots=\frac a{1+a}=\frac{\frac b{1-b}}{1+\frac b{1-b}}=b.$$

Answer 2

$a-a^2+a^3-a^4+\cdots=a+a(-a)+a(-a)^2+a(-a)^3+\cdots$ is also an infinite geometric series.

Note that $\displaystyle a=\frac{b}{1-b}=\frac{1}{1-b}-1$. When $\displaystyle -\frac{1}{2}<b<\frac{1}{2}$,

$$-\frac{1}{3}=\frac{1}{1-\frac{-1}{2}}-1<\frac{1}{1-b}-1<\frac{1}{1-\frac{1}{2}}-1=1$$

$$a-a^2+a^3-a^4+\cdots=\frac{a}{1-(-a)}=b$$