I'm trying to understand the Fourier series coefficients of the sum of two discrete-time periodic signals.
Consider two discrete-time periodic signals $x[n]$ and $y[n]$. $x[n]$ has period $N$, its Fourier series coefficients $a_k$. $y[n]$ has period $M$, its Fourier series coefficients $b_k$.
According to this problem solution (S10.10), the Fourier series coefficients $c_k$ of the summation signal $x[n] + y[n]$ can be represented by $a_k$, $b_k$ as follows:
$$c_k = \begin{cases} \frac{1}{N}a_{k/M} + \frac{1}{M}b_{k/N} &k \text{ a multiple of }M \text{ and } N\\ \frac{1}{N}a_{k/M}&k \text{ a multiple of }M\\ \frac{1}{M}b_{k/N}&k \text{ a multiple of }N\\ 0 &\text{otherwise} \end{cases}$$
I don't see where $\frac{1}{N}$ and $\frac{1}{M}$ come from. With a concrete example, say $N=3$ and $M=2$.
\begin{align} x[n] & = a_0 + 0 + a_1 e^{j\frac{4\pi}{6}n} + 0 + a_2 e^{j\frac{8\pi}{6}n} + 0\\ y[n] & = b_0 + 0 + 0 + b_1 e^{j\frac{6\pi}{6}n} + 0 + 0 \\ x[n]+y[n] &= c_0 + 0 + c_2 e^{j\frac{4\pi}{6}n} + c_3 e^{j\frac{6\pi}{6}n} + c_4 e^{j\frac{8\pi}{6}n} + 0 \end{align}
In this example,$c_0 = a_0 + b_0$, $c_1=0$, $c_2 = a_1$, $c_3 = b_1$, $c_4 = a_2$, $c_5 = 0$
To extrapolate, I think the correct formula to compute $c_k$ is
$$c_k = \begin{cases} a_{k/M} + b_{k/N} &k \text{ a multiple of }M \text{ and } N\\ a_{k/M}&k \text{ a multiple of }M\\ b_{k/N}&k \text{ a multiple of }N\\ 0 &\text{otherwise} \end{cases}$$
Also the formula doesn't make sense in special cases like $x[n] = y[n]$ and $y[n] = 0$.
If $y[n] = x[n]$, apparently $a_k = b_k$ and $c_k = 2a_k$. However, according to the formula above, $c_k = \frac{2}{N} a_{k/N}$ for $k$ a multiple of $N$ and $c_k = 0$ otherwise.
If $y[n] = 0$, $c_k = a_k$. But by the formula, $c_k = \frac{1}{N}a_{k/M}$. I don't see how the two representations of $c_k$ can be the same thing.
The original problem (P10.10)