Proof Verification: Lemma $2.4$ in Chapter $3$, Stein & Shakarchi's Fourier Analysis

Question

For each $N\ge 1$, we define the following two functions on $[-\pi,\pi]$; $$f_N(\theta) = \sum_{1\le |n|\le N} \frac{e^{in\theta}}{n} \quad \quad \widetilde{f_N}(\theta) = \sum_{-N \le n \le -1} \frac{e^{in\theta}}{n}$$ which are trigonometric polynomials of degree $N$. From these, we construct $P_N$ and $\widetilde{P_N}$ by defining $$P_N(\theta) = e^{i(2N)\theta} f_N(\theta)\quad\quad \widetilde{P_N}(\theta) = e^{i(2N)\theta} \widetilde{f_N}(\theta)$$ Furthermore, for a $2\pi$-periodic integrable function $f$ on the circle, we define the $N$th partial sum of its Fourier series by $S_N(f)= \sum_{|n|\le N} \hat f(n) e^{in\theta}$.

Lemma $2.4$: $$S_M(P_N) = \begin{cases}P_N & M\ge 3N\\ \widetilde{P_N} & M = 2N\\ 0 & M < N\end{cases}$$

Source: Chapter $3$, Stein & Shakarchi's Fourier Analysis

The proof seems to involve just computation, and so is not provided in the text. Here is my attempt.

My work: We have $$P_N(\theta) = \sum_{1\le |n| \le N} \frac{e^{i(2N + n) \theta}}{n}$$ and $$\hat{P_N}(m) = \frac{1}{2\pi} \int_{-\pi}^\pi P_N(\theta) e^{-im\theta}\, d\theta = \frac{1}{2\pi} \sum_{1\le |n| \le N} \int_{-\pi}^\pi \frac{e^{i(2N + n -m) \theta}}{n}\, d\theta$$ from which it is clear that $\hat{P_N}(m) \ne 0$ if and only if $m \in S: = \{N, \ldots, 2N-1\} \cup \{2N+1, \ldots, 3N\}$. In fact, $\hat{P_N}(m) = \frac{1}{m-2N}$ whenever $\hat{P_N}(m) \ne 0$. By definition of partial sums of the Fourier series of a function, $$S_M(P_N) = \sum_{|m| \le M} \hat P_N(m) e^{im\theta}$$ Note that for $\lambda\in \mathbb Z$, $$\frac{1}{2\pi}\int_{-\pi}^\pi e^{i\lambda\theta}\, d\theta = \begin{cases}1 & \lambda = 0\\ 0 & \lambda \ne 0\end{cases}$$

1. It follows that $S_M(P_N) = 0$ if $M < N$, for then $\hat{P_N}(m) = 0$ for all $|m|\le M$.

2. Now, consider $M \ge 3N$. Then, $$S_M(P_N) = \sum_{|m| \le M} \hat P_N(m) e^{im\theta} = \sum_{|m|\le M} \mathbf{1}_{S}(m) \frac{e^{im\theta}}{m-2N} = \sum_{1\le |n|\le N} \frac{e^{i(2N + n)\theta}}{n} = P_N(\theta)$$ by substituting $n = m- 2N$.

3. Lastly, suppose $M = 2N$. Then, $$S_M(P_N) = \sum_{|m| \le 2N} \hat P_N(m) e^{im\theta} = \sum_{N\le m \le 2N-1} \frac{e^{im\theta}}{m-2N} = \sum_{-N \le n\le -1} \frac{e^{i(2N+n) \theta}}{n} = \widetilde{P_N}(\theta)$$ by substituting $n = m - 2N$.

Please let me know if my proof is correct, and if there are other (possibly shorter) alternatives. Any other comments are also appreciated!