Uniform convergence of a sequence of functions given in the form of Fourier Series

Question

Let $f:[-\pi, \pi] \rightarrow \Bbb{R}$ be a continuous $2\pi$ periodic function whose Fourier series is given by
$\frac{a_0}2 + \displaystyle \sum_{k=1}^{\infty} \left( a_k \cos kt +b_k \sin kt\right) $ Let for each $n\in \Bbb{N} $,
$f_n(t) =\frac{a_0}2 + \displaystyle \sum_{k=1}^{n} \left( a_k \cos kt +b_k \sin kt\right) $,
and let $f_0$ denote the constant function $\frac{a_0}2$.Which of the following statements are true?
$(a) f_n \rightarrow f$ uniformly on $[-\pi, \pi] $
$(b)$ If $\sigma_n=\frac{f_0+f_1+...+f_n}{n+1}$, then $\sigma_n \rightarrow f $ uniformly on $[-\pi, \pi] $
$(c) \int_{-\pi} ^{\pi} |f_n(x) - f(x) |^2 dx \rightarrow 0,\text{as}\space n\rightarrow \infty$
$\bf{Try:}$
$(a) $ Each $f_n$ can be represented as a furies series. Now if we take $f(x) =\begin{cases} - 1 \quad x\in [-\pi, 0] \\ 1 \quad x\in [0,\pi] \end{cases}$
Then this $f$ can be represented as a Fourier Series. In that case each $f_n$ will be continuous but the pointwise limit $f$ is not. Hence cannot be Uniformly convergent.
$(b) $ I can't understand this
$(c) $ Since $f_n$ converges to $f$ pointwise, $|f_n - f|\le\epsilon$
for large $n$, then we have $\int_{-\pi} ^{\pi} |f_n(x) - f(x) |^2 dx \rightarrow 0$
as $n \rightarrow \infty$
$\bf{Need}$ : I want to learn if my thinking for $(a)$ is right or wrong. Whatever it be, I want to know if there is any theorem to solve this. Also I think the argument I have written in $(a) $ is not to the point. So, please correct me and tell me how to write the arguments properly mathematically. I'm unable to understand option $(b) $ so please describe this option.