What is Fourier series of $$f(x) = \sum_{n=1}^\infty \frac{\cos nx}{2^n}$$
Now, it was claimed that since $f(x)$ converge uniformly and:
$$f(x) = \sum_{n=1}^\infty \frac{e^{inx} + e^{i(-n)x}}{2^{n+1}} = \sum_{n\ne 1} \frac{e^{inx}}{2^{|n|+1}}$$
Then it must be that the Fourier coefficients are exactly the coefficients of the series.
Why is that?
The Fourier coefficient of $f$ are defined as taking the inner product of $f$ with a basic trigonometric polynomial. Because the series converges uniformly the order of integration and summation may be exchanged. Thus: $$<f,cos(nx)>=\int_{0}^{2\pi}f(x) \cos (nx)dx=\int_{0}^{2\pi}(\sum_{n=1}^{\infty}2^{-m}\cos(mx))\cos (nx)dx$$ $$=\sum_{m=1}^{\infty}\int_{0}^{2\pi}2^{-m}\cos(mx) \cos (nx)dx = \sum_{m=1}^\infty <2^{-m}\cos(mx),\cos(nx)>=2^{-n}\delta_{m,n}$$ A similar calculation shows that the coefficients of the sin functions are 0.