Question

Find the sum of the infinite series tan2 θ − tan4 θ + tan6 θ + . . . + (−1)n−1 tan2n θ + . . . whenever the series converges.

Answer 1

$\displaystyle\sum_{n=1}^\infty(-1)^{n-1}\tan^{2n}\theta=1-\sum_{n=0}^\infty(-\tan^2\theta)^n=1-\dfrac1{1+\tan^2\theta}$

where $|-\tan^2\theta|=\tan^2\theta<1$ in order that the series converges, which is to say the series converges for

$\tan^2\theta<1\implies|\tan\theta|<1\implies|\theta|<\dfrac\pi4$

We can simplify the sum a bit further, noting that $1+\tan^2\theta=\sec^2\theta$, giving

$\displaystyle\sum_{n=1}^\infty(-1)^{n-1}\tan^{2n}\theta=1-\cos^2\theta=\sin^2\theta$