If the Fourier coefficients satisfy $a_n=O(\frac{1}{n})$ and $f$ is continuous, then does $S_N(f)$ converges to $f$ uniformly? Suppose that $f$ is continuous over $\mathbb{T}$  with the following Fourier expansion 
$$f(x)\sim\sum_{n=-\infty}^{\infty}a_{n}e^{inx},$$
where
$$a_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)e^{-int}dt. $$
If for any $n$, $|a_n|\leq \frac{K}{n}$ for some constant $K$ that does not depend on $n$, then is it possible to show that the finite Fourier sum $S_N(f)$ converges to $f$ uniformly over $\mathbb{T}$ as $N\to \infty$?  
When $\sum |a_n|<+\infty$, we know that $S_N(f)$ converges to $f$ uniformly. The decay rate is sharp so that it is not absolutely summable.  
 A: Bonus: At the bottom there's a characterisation of the sequences in $L^1(\mathbb T)$ that serve as approximate identities for $C(\mathbb T)$.
It's true (this surprises me).
Say $\psi_\epsilon\in C^\infty_c(\mathbb R)$, $0\le\psi_\epsilon\le1$, with $\psi_\epsilon=1$ on $[-1,1]$ and $\psi_\epsilon=0$ on $(-\infty,-1-\epsilon]\cup[1+\epsilon,\infty)$.
Then $\psi_\epsilon=\hat F_\epsilon$ for some $F_\epsilon\in L^1$. Let $K_{n,\epsilon}$ be the trigonometric polynomial with $$\hat K_{n,\epsilon}(j)=\psi_\epsilon(j/n);$$
then it follows from by standard arguments that if $f\in C(\mathbb T)$ then $K_{n,\epsilon}*f\to f$ uniformly as $N\to\infty$ (for fixed $\epsilon$). (See Details below.) And it's clear that if $\hat f(j)=O(1/j)$ then $$||S_n(f)-K_n*f||\le c\sum_{j=n}^{(1+\epsilon)n}\frac1j\sim c\log(1+\epsilon)\sim c\epsilon$$for all $n$: hence $||S_nf-f||_\infty\le(c+1)\epsilon$ if $n$ is large enough.
Details Regarding why $K_{n,\epsilon}*f\to f$ uniformly:


Lemma. Suppose $F\in L^1(\mathbb R)$ and $\int F=1$. For $\lambda>0$ define $\delta_\lambda F(t)=\lambda F(\lambda t)$. If $f\in C(\mathbb R)$ is bounded and uniformly continuous then $\delta_\lambda F*f\to f$ uniformly as $\lambda\to\infty$.


Proof. Let $\epsilon>0$. Choose $d>0$ so $|f(x-t)-f(x)|<\epsilon$ if $|t|<d$. If $\lambda$ is large enough that $\int_{|t|>\lambda d}|F(t)|<\epsilon$ then $$f(x)-\delta_\lambda F*f(x)
=\int(f(x)-f(x-t))\delta_\lambda F(t)=\int_{|t|<d}+\int_{|t|>d}=I+II.$$
Now $$||I||<\epsilon||\delta_\lambda F||_1=\epsilon||F||_1,$$
while $$||II||\le2||f||_\infty\int_{|t|>d}|\delta_\lambda F(t)|
=2||f||_\infty\int_{|t|>\lambda d}|F(t)|<2||f||_\infty\epsilon.$$
Now let $\psi_\epsilon$ be as above, and choose $F_\epsilon\in L^1(\mathbb R)$ with $\psi_\epsilon=\hat F_\epsilon$. Note that $\int F_\epsilon=\psi_\epsilon(0)=1$. Define $$K_{n,\epsilon}(t)=2\pi\sum_{k\in\mathbb Z}\delta_nF_\epsilon(t+2\pi k).$$
Now if $\phi$ is bounded and has period $2\pi$ it follows that $$\frac1{2\pi}\int_0^{2\pi}K_{n,\epsilon}(t)\phi(t)
=\int_{-\infty}^\infty \delta_n F_\epsilon(t)\phi(t).$$This has two consequences:
First, if $j\in\mathbb Z$ then $$\hat K_{n,\epsilon}(j)=\widehat{\delta_n F_\epsilon}(j)=\hat F_\epsilon(j/n),$$as advertised. Second, if $f$ is continuous and $2\pi$-periodic then $$K_{n,\epsilon}*f=(\delta_n F_\epsilon)*f\to f$$uniformly. (Where of course the first convolution  in the equation is convolution on $\mathbb T$ and the second is convolution on $\mathbb R$.)
What do we learn from all this? Here's the take-away. Say $D_n$ is the Dirichlet kernel.


Moral: The reason $S_nf$ does not converge to $f$ for every continuous function $f$ is that $||D_n||_1$ is unbounded; the reason for this is that $\chi_{[-1,1]}$ is not the Fourier transform of an $L^1$ function. But if we smooth out the discontinuity in $\chi_{[-1,1]}$ then all the $L_1$ norms stay bounded and everything works.


Chuckle: I just realized one can prove that the $L^1$ norm really is all that matters here:


Theorem. Suppose $(K_n)\subset L^1(\mathbb T)$. Then $K_n*f\to f$ uniformly for all $f\in C(\mathbb T)$ if and only if $\hat K_n(j)\to 1$ for every $j$ and $||K_n||_1$ is bounded.


Proof: If $e_k(t)=e^{ikt}$ then $K_n*e_k=\hat K_n(k) e_k$, so if $K_n*f\to f$ for every $f$ then $\hat K_n(k)\to1$. And if $||K_n||_1$ is unbounded then Uniform Boundedness shows that there exists $f\in C(\mathbb T)$ such that $K_n*f(0)$ is unbounded.
Conversely, suppose that $||K_n||_1\le c$ and $\hat K_n(k)\to 1$ for every $k$. Suppose $f\in C(\mathbb T)$. Let $\epsilon>0$. Choose a trigonoetric polynomial $P$ with $||f-P||_\infty<\epsilon$. Then $$||K_n*f-K_n*P||_\infty\le c\epsilon$$for every $n$ and $K_n*P\to P$ uniformly, hence $$||f-K_n*f||_\infty<(c+2)\epsilon$$if $n$ is large enough.
