= Solution
With $\widehat f(k)=(2\pi)^{-1}\int_{-\pi}^{\pi}f(s)e^{-iks}ds$, the classical <Fourier partial sum> is $s_N^f(t)=\sum_{|k|\leq N}\widehat f(k)e^{ikt}$. Its <Cesàro mean> is
$$
\sigma_N^f=\frac1{N+1}\sum_{n=0}^Ns_n^f=\sum_{|k|\leq N}\left(1-\frac{|k|}{N+1}\right)\widehat f(k)e^{ikt}.
$$
For the uniform <Fejér theorem>, extend the continuous endpoint-matching $f$ periodically. The given <Fejér kernel> is nonnegative, even and has total integral $\pi$. Thus
$$
\sigma_N^f(t)-f(t)=\frac1\pi\int_{-\pi}^{\pi}[f(t-u)-f(t)]F_N(u)du.
$$
Given $\varepsilon>0$, uniform <continuity> supplies $\delta>0$ so the bracket has modulus less than $\varepsilon$ for $|u|<\delta$, uniformly in $t$. That part contributes at most $\varepsilon$. On the remainder, $F_N(u)\leq[2(N+1)\sin^2(\delta/2)]^{-1}$, and the bracket is at most $2\|f\|_\infty$. Hence
$$
\|\sigma_N^f-f\|_\infty\leq\varepsilon+\frac{2\|f\|_\infty}{(N+1)\sin^2(\delta/2)}\longrightarrow\varepsilon.
$$
Letting $\varepsilon$ decrease to zero proves \b[<uniform convergence> of the <Fejér sums>].
The later assertions attached to the printed factorial-frequency example are inconsistent. The following calculations resolve them while keeping the Fourier definition just given. Put $a_j=e^{i\sqrt{j+1}}-e^{i\sqrt j}$ and consider its formal trigonometric <series>. For $t=\pi p/q$, there is $J_0$ such that $\cos(j!t)=1$ for every $j\geq J_0$, because $2q$ divides $j!$. At $N_J=(J+1)!-1$, its ordinary partial sum therefore telescopes to
$$
s_{N_J}(t)=C(t)+e^{i\sqrt{J+1}}.
$$
The exponential has subsequences tending to $1$ and $-1$: take $J+1$ to be integers nearest $(2\pi m)^2$ and $((2m+1)\pi)^2$. Thus \b[the ordinary sums diverge at every $t\in\pi\mathbb Q$].
But their genuine Fourier <Cesàro means> also diverge there. Indeed,
$$
\sigma_{N_J}(t)-s_{N_J}(t)=-\frac1{(J+1)!}\sum_{j=0}^Jj!a_j\cos(j!t).
$$
Since $|a_j|\leq2$ and $\sum_{j=0}^Jj!\leq2J!$ for $J\geq3$, its absolute value is at most $4/(J+1)\to0$, uniformly in $t$. Therefore
$$
\boxed{\sigma_{N_J}(t)=C(t)+e^{i\sqrt{J+1}}+o(1),}
$$
which has the same incompatible limit points. This is an instance of <Cesàro sums at factorial cutoffs track ordinary sums>. \b[The printed Cesàro-convergence claim is false.] Averaging the sparse <series> by its index $j$ would be a different operation from Fourier Cesàro summation by frequency.
There is a further issue: \b[no $L^1$ function has exactly this <Fourier series>]. Here is a direct proof using <strongly lacunary polynomial L1-L2 bound>. For a <trigonometric polynomial> $p$ supported on $\{\pm j!:j\geq3\}$, consecutive positive frequencies have ratio at least four. Every zero sum of four signed frequencies pairs into cancelling equal frequencies: a largest unpaired frequency would exceed the sum of the other three. Expanding the fourth moment with normalized measure $dt/(2\pi)$ and bounding the three possible pairings gives
$$
\|p\|_4^4\leq3\|p\|_2^4.
$$
The <Holder inequality> gives $\|p\|_2\leq\|p\|_1^{1/3}\|p\|_4^{2/3}$, hence $\|p\|_2\leq\sqrt3\|p\|_1$.
If the asserted $f\in L^1$ existed, positivity and unit mass of the summation kernel would bound $\|\sigma_N^f\|_1\leq\|f\|_1$. Subtract the finitely many modes from $j<3$; their norms stay uniformly bounded, and the remainder is a <polynomial> of the type above. Its $L^2$ norms would therefore be uniformly bounded. <Orthogonality> then forces $\sum_{j\geq3}|a_j|^2<\infty$, by fixing any finite coefficient block and letting $N\to\infty$. However,
$$
|a_j|^2=4\sin^2\left(\frac{\sqrt{j+1}-\sqrt j}{2}\right)\sim\frac1{4j},
$$
so that sum diverges. This contradiction disproves the stated existence premise.
Even considered only formally, these coefficients rule out an $L^2$ function by <Bessel's inequality>, and hence rule out a bounded function on the finite interval. Thus a bounded continuous representative cannot satisfy the asserted <series> either. \b[The asserted $L^1$ function does not exist.]
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