## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{1+2r}{1-r^2}$ for $|r^2| \lt 1$ or, $|r| \lt 1$
Since, we have $s_n=1+2r+r^2+2r^3+r^4+2r^5+.....$ This can be re-written as: $s_n=(1+r^2+r^4+...+r^{2n})+(2r+2r^3+2r^5+...+2r^{2n+1})$ Now, $\lim\limits_{n \to \infty} s_n=\dfrac{1}{1-r^2}+\dfrac{2r}{1-r^2}=\dfrac{1+2r}{1-r^2}$ for $|r^2| \lt 1$ or, $|r| \lt 1$