Consider:
$$f_{n,N}(z) = \sin(Nx) \sum_{k = 1}^n \frac{\sin(kx)}{k}$$
Now consider
$$\sum_k \frac{1}{k^2} f_{2^{k^3}, 2^{k^3 - 1}}(z)$$
Now for $x = \pi / (4n)$ and $N = 2n$ we get that
$$\sin(\pi/4) \sum_1^n \frac{1}{k} > \frac{1}{\sqrt{2}} \log n$$
So we have for some $x$ that
$$|s_{n_k + 1} - s_{n_k}| \geq \frac{1}{\sqrt{2}} \frac{1}{k^2} \log n_k$$
So we cannot have uniform convergence. I believe this is due to Hugo Steinhaus.
I hope I didn't make a mistake, but it is along these lines, I can correct it if I made an error.
This is a theorem of Bernstein, found, for example, in Katznelson's Introduction to Harmonic Analysis.
A simple-minded approach would be to shift the variable in $$c_n=\frac{1}{2\pi}\int_0^{2\pi} f(x)e^{-inx}\,dx\tag1$$ to $y=x+\pi/n$, thus obtaining
$$c_n=\frac{1}{2\pi}\int_{0}^{2\pi} f_{\pi/n}(y)e^{-iny+\pi i}\,dy\tag2$$
where I write $f_h(x)=f(x-h)$. Then add (1) and (2):
$$2c_n=\frac{1}{2\pi}\int_0^{2\pi} (f(x)-f_{\pi/n}(x))e^{-inx}\,dx\tag3$$
and weep in despair: the estimate is only $|c_n|=O(n^{-\alpha})$...
The thing is, estimating $|c_n|$ by maximum of integrand in (3) and then summing the estimates is just too crude. Parseval's identity is more precise.
So, let's try Parseval. Since the $n$th Fourier coefficient of $f_h-f$ is $(e^{-inh}-1)\widehat{f}(n)$, we have
$$ \sum_n |e^{-inh}-1|^2|\widehat{f}(n)|^2 \le Ch^{2\alpha} \tag4$$
The inequality (4) isn't worth much unless we can bound $ |e^{-inh}-1|$ from below. There is no uniform bound for all $n$, but if we focus on some dyadic range $2^k\le |n|< 2^{k+1}$, then choosing $h=\frac{2\pi}{3}\cdot 2^{-k}$ gives good result: $\frac{2\pi}{3}\le |nh|< \frac{4\pi}{3}$, which keeps $e^{-inh}$ far away from $1$.
So,
$$ \sum_{n=2^k}^{2^{k+1}-1} |\widehat{f}(n)|^2 \le \widetilde{C} 2^{-2k\alpha} \tag5$$
and the rest is a mopping-up operation: Cauchy-Schwarz turns (5) into
$$ \sum_{n=2^k}^{2^{k+1}-1} |\widehat{f}(n)| \le 2^{k/2}\cdot\widetilde{C} 2^{-k\alpha} \tag6$$
which sums up.
Best Answer
Let $$ D_N(t)=\frac{1}{2\,\pi}\,\frac{\sin(N+1/2)t}{\sin(t/2)} $$ be the Dirichlet kernel, and let $$ S_N(f;x)=\sum_{k=-N}^N\hat f(k)e^{ikx}=\int_{-\pi}^{\pi}D_N(t)f(x-t)\,dt $$ be the $N$-th partial sum of the Fourier series of $f$. Then, for all $x\in[-\pi,\pi]$ $$ S_N(f,x)-f(x)=\int_{-\pi}^{\pi}D_N(t)(f(x-t)-f(x))\,dt=\int_{-\pi}^{\pi}\sin\frac{(2\,N+1)t}{2}h(t)\,dt $$ where $$ h(t)=\frac{f(x-t)-f(x)}{\sin(t/2)}\;. $$ Since $f$ is Hölder continuous of order $\alpha$, it follows that $|h(t)|\le Ct^{-1+\alpha}$ for some constant $C>0$. In particular, $h$ is integrable on $[-\pi,\pi]$. The proof is finished applying the Riemann-Lebesgue lemma.
It should be noted that your problem is a particular case of Dini's convergence criterion.