Prove that $\sum_{k=0}^{n-1} \frac{\cos(k\cos^{-1}x)}{x^k}=\frac {\sin(n\cos^{-1}x)}{x^{n-1}\sqrt[2]{1-x^2}} $

Question

By considering $\sum_{k=0}^{n-1} (1+i\tan\theta)^k,$ how do I prove that

$$\sum_{k=0}^{n-1} \frac{\cos(k\cos^{-1}x)}{x^k}=\frac {\sin(n\cos^{-1}x)}{x^{n-1}\sqrt[2]{1-x^2}}$$ for $0<x<1$

I am completely clueless about how to approach this problem

Answer 1

Note that

$$\sum_{k=0}^{n-1}(1+i\tan(\theta))^k=\sum_{k=0}^{n-1}\frac{[\cos(\theta)+i\sin(\theta)]^k}{\cos^k(\theta)}=\sum_{k=0}^{n-1}\frac{\cos(k\theta)+i\sin(k\theta)}{\cos^k(\theta)}$$

The first sum is given by geometric series

$$\sum_{k=0}^{n-1}(1+i\tan(\theta))^k=\frac{(1+i\tan(\theta))^n-1}{1+i\tan(\theta)-1}=\frac{\cos(n\theta)-\cos^n(\theta)+i\sin(n\theta)}{i\cos^n(\theta)\tan(\theta)}$$

Taking real parts gives

$$\sum_{k=0}^{n-1}\frac{\cos(k\theta)}{\cos^k(\theta)}=\frac{\sin(n\theta)}{\cos^{n-1}(\theta)\sin(\theta)}$$

Now just substitute $x=\cos(\theta)$ and you will have your result.