$\sum _{n=1}^{\infty } \frac{\left(-1\right)^{n+1}\cos^n\left(x\right)}{2^{n-1}}$ converges to. Consider the series
$$\sum _{n=1}^{\infty } \frac{\left(-1\right)^{n+1}\cos^n\left(x\right)}{2^{n-1}}$$
I know that the series converges absolutely by comparison test. But finding difficulty to find where it converges to.
 A: Note that\begin{align}\sum_{n=1}^\infty\frac{(-1)^{n+1}\cos^n(x)}{2^{n-1}}&=-2\sum_{n=1}^\infty\left(\frac{-\cos(x)}2\right)^n\\&=-2\frac{-\frac{\cos x}2}{1+\frac{\cos x}2}\\&=\frac{\cos x}{1+\frac{\cos x}2}.\end{align}
A: If $|r|< 1$, we have$$\sum_{n=0}^{\infty}(-1)^{n} r^{n}=\frac{1}{1+r}.$$
Does this help?
A: Starting from first step of the very nice user @José Carlos Santos that is equivalent to
$$\left(- 2 \sum_{n=1}^{\infty} \left(- \frac{\cos{\left(x \right)}}{2}\right)^{n}\right)\tag 1$$
the term
$$-\frac{\cos{\left(x \right)}}{2}\leq\frac{\left|{\cos{\left(x \right)}}\right|}{2} < 1 \tag 2$$
and $$\sum_{n=1}^{\infty} \left(- \frac{\cos{\left(x \right)}}{2}\right)^{n}=-\frac{\cos{\left(x \right)}}{2 \left(\frac{\cos{\left(x \right)}}{2} + 1\right)} \tag 3$$ is an infinite geometric series and the sum it is convergent for the $(2)$.
Hence, for the $(1)$
$$-2\cdot \left(-\frac{\cos{\left(x \right)}}{2 \left(\frac{\cos{\left(x \right)}}{2} + 1\right)}\right)=\frac{2 \cos{\left(x \right)}}{\cos{\left(x \right)} + 2}$$
that is the sum of the series.
