This question is related to the construction of a continuous function with a divergent Fourier series (at $0$), as done in Stein and Shakarchi's Fourier Analysis. The construction uses Lemma $2.4$, which I have posted here. For completeness, I shall provide all the necessary background.
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 $N$. It is shown that $|\widetilde{f_N}(0)| \ge c\log N$, and $f_N$ is uniformly bounded in $N$ and $\theta$. 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 $2.4$: $$S_M(P_N) = \begin{cases}P_N & M\ge 3N\\ \widetilde{P_N} & M = 2N\\ 0 & M < N\end{cases}$$
Finally, we take a convergent series of positive terms $\sum \alpha_k$ and a sequence of integers $\{N_k\}$ satisfying $N_{k+1} > 3N_k$ for all $k$, and $\alpha_k\log N_k \xrightarrow{k\to\infty} \infty$. For example, $\alpha_k = \frac{1}{k^2}$ and $N_k = 3^{2^k}$ would suffice. The required continuous function $f$, with divergent Fourier series at $0$, is defined as $$f(\theta) = \sum_{k=1}^\infty \alpha_k P_{N_k}(\theta)$$
Using the Weierstrass M-test, the convergence of $\sum_{k=1}^\infty \alpha_k P_{N_k}(\theta)$ due to convergence of $\alpha_k$ and uniform boundedness of $P_N$ (note that $|P_N| = |f_N|$), and the continuity of $\{P_{N_k}\}$, we conclude that the series which defines $f$, converges uniformly (and absolutely) to a continuous periodic function.
Then, the author goes on to say that by the aforementioned lemma, we have $$|S_{2N_m}(f)(0)| \ge c\alpha_m\log N_m + \mathcal O(1) \xrightarrow{m\to\infty} \infty$$
Question: How does one conclude $$|S_{2N_m}(f)(0)| \ge c\alpha_m\log N_m + \mathcal O(1)$$
My thoughts: $$\begin{align*}
S_{2N_m}(f)(0) = \sum_{|n| \le 2N_m} \hat f(n) &= \sum_{|n| \le 2N_m} \sum_{k=1}^\infty \alpha_k \hat{P_{N_k}}(n)\\ &= \sum_{k=1}^\infty \sum_{|n| \le 2N_m} \alpha_k \hat{P_{N_k}}(n)\\ &= \sum_{k=1}^\infty\alpha_k S_{2N_m} (P_{N_k})(0)
\end{align*}$$
since limits and finite sums always commute, and
$$\hat f(n) = \frac{1}{2\pi}\int_{-\pi}^\pi \sum_{k=1}^\infty \alpha_k P_{N_k}(\theta) e^{-in\theta}\, d\theta = \sum_{k=1}^\infty \frac{1}{2\pi}\int_{-\pi}^\pi \alpha_k P_{N_k}(\theta) e^{-in\theta}\, d\theta = \sum_{k=1}^\infty \alpha_k \hat{P_{N_k}}(n)$$
where the exchange of the integral and infinite sum is justified by the uniform convergence of $\sum_{k=1}^\infty \alpha_k P_{N_k}(\theta)$, which follows from the Weierstrass M-test.
Best Answer
I figured it out on my own, so I am posting an answer for completeness's sake. $$\begin{align*} S_{2N_m}(f)(0) = \sum_{|n| \le 2N_m} \hat f(n) &= \sum_{|n| \le 2N_m} \sum_{k=1}^\infty \alpha_k \hat{P_{N_k}}(n)\\ &= \sum_{k=1}^\infty \sum_{|n| \le 2N_m} \alpha_k \hat{P_{N_k}}(n)\\ &= \sum_{k=1}^\infty\alpha_k S_{2N_m} (P_{N_k})(0)\\ &= \sum_{k=1}^{m} \alpha_k S_{2N_m} (P_{N_k})(0) \\ &= \alpha_m S_{2N_m} (P_{N_m})(0) + \sum_{k=1}^{m-1} \alpha_k S_{2N_m} (P_{N_k})(0)\\ &= \alpha_m \widetilde{P_{N_m}}(0) + \sum_{k=1}^{m-1} \alpha_k P_{N_k}(0) \end{align*}$$ We have $$\left| \alpha_m \widetilde{P_{N_m}}(0) \right| = \alpha_m |\widetilde{f_{N_m}}(0)| \ge c \alpha_m \log N_m$$ and $$\left|\sum_{k=1}^{m-1} \alpha_k P_{N_k}(0) \right| \le \sup_{n\ge 1} |f_n(0)| \sum_{k=1}^\infty \alpha_k < \infty$$ so $$\begin{align*}|S_{2N_m}(f)(0)| &= \left|\alpha_m \widetilde{P_{N_m}}(0) + \sum_{k=1}^{m-1} \alpha_k P_{N_k}(0) \right| \\ &\ge \left|\alpha_m \widetilde{P_{N_m}}(0) \right| - \left|\sum_{k=1}^{m-1} \alpha_k P_{N_k}(0)\right|\\ &\ge c\alpha_m \log N_m - \sup_{n\ge 1} |f_n(0)| \sum_{k=1}^\infty \alpha_k \end{align*}$$ giving $$|S_{2N_m}(f)(0)| \ge c\alpha_m\log N_m + \mathcal O(1)$$
Note: To get $$\sum_{k=1}^\infty\alpha_k S_{2N_m} (P_{N_k})(0) = \sum_{k=1}^{m} \alpha_k S_{2N_m} (P_{N_k})(0)$$ we have used $N_{k+1} > 3N_k$ for all $k$, and Lemma $2.4$ as stated above.