Question : for what values of a does the following series converge ?
\[S=1+3a+a^2+3a^3+a^4+3a^5+a^6+\ldots\]
Solution :
Let Given series as \[S=1+3a+a^2+3a^3+a^4+3a^5+a^6+\ldots\] This series can be factored as \[S\ =\ (1+a^2+a^4+\ldots)+3a\left(1+a^2+a^4+..\right)\] \[S=\left(1+3a\right)\left(1+a^2+a^4+\ldots\right)\] Since 1+3a is finite for all b and \[1+a^2+a^4+\ldots\] is a geometric series with common ratio a^2. This infinite geometric converges if \[\left|a^2\right|<1\] and it’s sum is \[\frac{1}{1-a^2}\] Hence the sum of the series \[S=\left(1+3a\right)\left(\frac{1}{1-a^2}\right)\ \ \ if\ \ \left|a^2\right|<1\] \[S=\left(1+3a\right)\left(\frac{1}{1-a^2}\right)\ \ \ if\ \ \left|a\right|^2<1\] \[S=\left(1+3a\right)\left(\frac{1}{1-a^2}\right)\ \ \ if\ \ \left|a\right|<1\] Thus the given series converges for all values of a in the interval (-1, 1) and it’s is given as \[\left(1+3a\right)\left(\frac{1}{1-a^2}\right)\]