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-\frac1{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|<\frac\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`