Coefficients of the Fourier Series expansion of physical signals having finite power

Question 2.7. Show that for all periodic physical signals that have finite power, the coefficients of the Fourier series expansion $x_n$ tends to 0 as $n \to \infty$

I have computed $|x_n|^2$ by multiplying $x_n$ with $x_n^*$. However, as per my calculations, $|x_n|^2 = P_x$ i.e. power of $x(t)$, which is a constant and independent of $n$. I know that this is fundamentally incorrect because $|x_n|$ should vary with $n$. If anyone can guide me about where I went wrong that would be very helpful. Thanks and regards.

MY SOLUTION:

Please refer to the solution from the half of the first page.

Answer

A finite power periodic signal will have the following property:

$$P_x = \frac{1}{T_0} \int_{0}^{T_0} |x(t)|^2 dt < \infty \tag{1}$$

where $P_x$ is the power averaged over a period of the signal. Then from Parseval's theorem, the total power can also be shown to be (for a real signal I'm assuming):

$$P_x = |a_0|^2+2 \sum_{n=1}^{\infty} |a_n|^2 < \infty \tag{2}$$

where the $a_n$'s are the continuous-time Fourier series coefficients for $x(t)$.

From the theory of power series (calculus), it's known that in order for (2) to converge (sum being less than infinity) a necessary condition is

$$\lim_{n \to \infty} |a_n|^2 = 0 \tag{3}$$

(3) implies therefore that, for finite power periodic signal's CTFS coefficients $a_n$ also goes to zero as $n$ goes to infinity.